{"id":119094,"date":"2022-05-02T18:07:07","date_gmt":"2022-05-02T12:37:07","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119094"},"modified":"2022-05-02T18:07:07","modified_gmt":"2022-05-02T12:37:07","slug":"chapter-5-arithmetic-progressions-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-5-arithmetic-progressions-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 5 – Arithmetic Progressions Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 5.1<\/h2>\n

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
\n(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.<\/h2>\n

Solution:
\nWe can write the given condition as;
\nTaxi fare for 1 km = 15
\nTaxi fare for first 2 kms = 15+8 = 23
\nTaxi fare for first 3 kms = 23+8 = 31
\nTaxi fare for first 4 kms = 31+8 = 39
\nAnd so on\u2026\u2026
\nThus, 15, 23, 31, 39 \u2026 forms an A.P. because every next term is 8 more than the preceding term.<\/h3>\n

(ii) The amount of air present in a cylinder when a vacuum pump removes 1\/4\u00a0of the air remaining in the cylinder at a time.<\/h2>\n

Solution:
\nLet the volume of air in a cylinder, initially, be\u00a0V\u00a0litres.
\nIn each stroke, the vacuum pump removes\u00a01\/4th\u00a0of air\u00a0remaining in the cylinder at a time. Or we can say, after every stroke, 1-1\/4 = 3\/4th part of air will remain.
\nTherefore, volumes will be\u00a0V, 3V\/4 , (3V\/4)2\u00a0, (3V\/4)3\u2026and so on
\nClearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.<\/h3>\n

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.<\/h2>\n

Solution:
\nWe can write the given condition as;
\nCost of digging a well for first metre = Rs.150
\nCost of digging a well for first 2 metres = Rs.150+50 = Rs.200
\nCost of digging a well for first 3 metres = Rs.200+50 = Rs.250
\nCost of digging a well for first 4 metres =Rs.250+50 = Rs.300
\nAnd so on..
\nClearly, 150, 200, 250, 300 \u2026 forms an A.P. with a common difference of 50 between each term.<\/h3>\n

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.<\/h2>\n

Solution:
\nWe know that if Rs. P\u00a0is deposited at\u00a0r% compound interest per annum for n years, the amount of money will be:
\nP(1+r\/100)n
\nTherefore, after each year, the amount of money will be;
\n10000(1+8\/100), 10000(1+8\/100)2, 10000(1+8\/100)3\u2026\u2026
\nClearly, the terms of this series do not have the common difference between them. Therefore, this is\u00a0not an A.P.<\/h3>\n

2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:
\n(i)\u00a0a\u00a0= 10,\u00a0d\u00a0= 10
\n(ii)\u00a0a\u00a0= -2,\u00a0d\u00a0= 0
\n(iii)\u00a0a\u00a0= 4,\u00a0d\u00a0= \u2013 3
\n(iv)\u00a0a\u00a0= -1\u00a0d\u00a0= 1\/2
\n(v)\u00a0a\u00a0= \u2013 1.25,\u00a0d\u00a0= \u2013 0.25<\/h2>\n

Solutions:
\n(i)\u00a0a\u00a0= 10,\u00a0d\u00a0= 10
\nLet us consider, the Arithmetic Progression series be\u00a0a1,\u00a0a2,\u00a0a3,\u00a0a4,\u00a0a5\u00a0\u2026
\na1\u00a0=\u00a0a\u00a0= 10
\na2\u00a0=\u00a0a1+d\u00a0= 10+10 = 20
\na3\u00a0=\u00a0a2+d\u00a0= 20+10 = 30
\na4\u00a0=\u00a0a3+d\u00a0= 30+10 = 40
\na5\u00a0=\u00a0a4+d\u00a0= 40+10 = 50
\nAnd so on\u2026
\nTherefore, the A.P. series will be 10, 20, 30, 40, 50 \u2026
\nAnd First four terms of this A.P. will be 10, 20, 30, and 40.<\/h3>\n

(ii)\u00a0a\u00a0= \u2013 2,\u00a0d\u00a0= 0
\nLet us consider, the Arithmetic Progression series be\u00a0a1,\u00a0a2,\u00a0a3,\u00a0a4,\u00a0a5\u00a0\u2026
\na1\u00a0=\u00a0a\u00a0= -2
\na2\u00a0=\u00a0a1+d\u00a0= \u2013 2+0 = \u2013 2
\na3\u00a0=\u00a0a2+d = \u2013 2+0 = \u2013 2
\na4\u00a0=\u00a0a3+d\u00a0= \u2013 2+0 = \u2013 2
\nTherefore, the A.P. series will be \u2013 2, \u2013 2, \u2013 2, \u2013 2 \u2026
\nAnd, First four terms of this A.P. will be \u2013 2, \u2013 2, \u2013 2 and \u2013 2.<\/h3>\n

(iii)\u00a0a\u00a0= 4,\u00a0d\u00a0= \u2013 3
\nLet us consider, the Arithmetic Progression series be\u00a0a1,\u00a0a2,\u00a0a3,\u00a0a4,\u00a0a5\u00a0\u2026
\na1\u00a0=\u00a0a\u00a0= 4
\na2\u00a0=\u00a0a1+d\u00a0= 4-3 = 1
\na3\u00a0=\u00a0a2+d\u00a0= 1-3 = \u2013 2
\na4\u00a0=\u00a0a3+d\u00a0= -2-3 = \u2013 5
\nTherefore, the A.P. series will be 4, 1, \u2013 2 \u2013 5 \u2026
\nAnd, first four terms of this A.P. will be 4, 1, \u2013 2 and \u2013 5.<\/h3>\n

(iv)\u00a0a\u00a0= \u2013 1,\u00a0d\u00a0= 1\/2
\nLet us consider, the Arithmetic Progression series be\u00a0a1,\u00a0a2,\u00a0a3,\u00a0a4,\u00a0a5\u00a0\u2026
\na2\u00a0=\u00a0a1+d\u00a0= -1+1\/2 = -1\/2
\na3\u00a0=\u00a0a2+d\u00a0= -1\/2+1\/2 = 0
\na4\u00a0=\u00a0a3+d\u00a0= 0+1\/2 = 1\/2
\nThus, the A.P. series will be-1, -1\/2, 0, 1\/2
\nAnd First four terms of this A.P. will be -1, -1\/2, 0 and 1\/2.<\/h3>\n

(v)\u00a0a\u00a0= \u2013 1.25,\u00a0d\u00a0= \u2013 0.25
\nLet us consider, the Arithmetic Progression series be\u00a0a1,\u00a0a2,\u00a0a3,\u00a0a4,\u00a0a5\u00a0\u2026
\na1\u00a0=\u00a0a\u00a0= \u2013 1.25
\na2\u00a0=\u00a0a1\u00a0+\u00a0d\u00a0= \u2013 1.25-0.25 = \u2013 1.50
\na3\u00a0=\u00a0a2\u00a0+\u00a0d\u00a0= \u2013 1.50-0.25 = \u2013 1.75
\na4\u00a0=\u00a0a3\u00a0+\u00a0d\u00a0= \u2013 1.75-0.25 = \u2013 2.00
\nTherefore, the A.P series will be 1.25, \u2013 1.50, \u2013 1.75, \u2013 2.00 \u2026\u2026..
\nAnd first four terms of this A.P. will be \u2013 1.25, \u2013 1.50, \u2013 1.75 and \u2013 2.00.<\/h3>\n

3. For the following A.P.s, write the first term and the common difference.
\n(i) 3, 1, \u2013 1, \u2013 3 \u2026
\n(ii) -5, \u2013 1, 3, 7 \u2026
\n(iii) 1\/3, 5\/3, 9\/3, 13\/3 \u2026.
\n(iv) 0.6, 1.7, 2.8, 3.9 \u2026<\/h2>\n

Solutions
\n(i) Given series,
\n3, 1, \u2013 1, \u2013 3 \u2026
\nFirst term,\u00a0a\u00a0= 3
\nCommon difference,\u00a0d\u00a0= Second term \u2013 First term
\n\u21d2\u00a0 1 \u2013 3 = -2
\n\u21d2\u00a0 d = -2<\/h3>\n

(ii) Given series, \u2013 5, \u2013 1, 3, 7 \u2026
\nFirst term,\u00a0a\u00a0= -5
\nCommon difference,\u00a0d\u00a0= Second term \u2013 First term
\n\u21d2\u00a0( \u2013 1)-( \u2013 5) = \u2013 1+5 = 4<\/h3>\n

(iii) Given series, 1\/3, 5\/3, 9\/3, 13\/3 \u2026.
\nFirst term,\u00a0a\u00a0= 1\/3
\nCommon difference,\u00a0d\u00a0= Second term \u2013 First term
\n\u21d2\u00a05\/3 \u2013 1\/3 = 4\/3<\/h3>\n

(iv) Given series, 0.6, 1.7, 2.8, 3.9 \u2026
\nFirst term,\u00a0a\u00a0= 0.6
\nCommon difference,\u00a0d\u00a0= Second term \u2013 First term
\n\u21d2\u00a01.7 \u2013 0.6
\n\u21d2\u00a01.1<\/h3>\n

4. Which of the following are APs? If they form an A.P. find the common difference\u00a0d\u00a0and write three more terms.
\n(i) 2, 4, 8, 16 \u2026
\n(ii) 2, 5\/2, 3, 7\/2 \u2026.
\n(iii) -1.2, -3.2, -5.2, -7.2 \u2026
\n(iv) -10, \u2013 6, \u2013 2, 2 \u2026
\n(v) 3, 3 +\u00a0\u221a2, 3\u00a0+ 2\u221a2, 3\u00a0+ 3\u221a2
\n(vi) 0.2, 0.22, 0.222, 0.2222 \u2026.
\n(vii) 0, \u2013 4, \u2013 8, \u2013 12 \u2026
\n(viii) -1\/2, -1\/2, -1\/2, -1\/2 \u2026.
\n(ix) 1, 3, 9, 27 \u2026
\n(x)\u00a0a, 2a, 3a, 4a\u00a0\u2026
\n(xi)\u00a0a,\u00a0a2,\u00a0a3,\u00a0a4\u00a0\u2026
\n(xii) \u221a2, \u221a8, \u221a18, \u221a32\u00a0\u2026
\n(xiii) \u221a3, \u221a6, \u221a9, \u221a12\u00a0\u2026
\n(xiv) 12, 32, 52, 72\u00a0\u2026
\n(xv) 12, 52, 72, 73\u00a0\u2026<\/h2>\n

Solution
\n(i) Given to us,
\n2, 4, 8, 16 \u2026
\nHere, the common difference is;
\na2\u00a0\u2013\u00a0a1\u00a0= 4 \u2013 2 = 2
\na3\u00a0\u2013\u00a0a2\u00a0= 8 \u2013 4 = 4
\na4\u00a0\u2013\u00a0a3\u00a0= 16 \u2013 8 = 8
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not the same every time.
\nTherefore, the given series are not forming an A.P.<\/h3>\n

(ii) Given, 2, 5\/2, 3, 7\/2 \u2026.
\nHere,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a05\/2-2 = 1\/2
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a03-5\/2 = 1\/2
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a07\/2-3 = 1\/2
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0= 1\/2\u00a0and the given series are in A.P.
\nThe next three terms are;
\na5\u00a0= 7\/2+1\/2 = 4
\na6\u00a0= 4\u00a0+1\/2 = 9\/2
\na7\u00a0= 9\/2\u00a0+1\/2 = 5<\/h3>\n

(iii)\u00a0Given, -1.2, \u2013 3.2, -5.2, -7.2 \u2026
\nHere,
\na2\u00a0\u2013\u00a0a1\u00a0= (-3.2)-(-1.2) = -2
\na3\u00a0\u2013\u00a0a2\u00a0= (-5.2)-(-3.2) = -2
\na4\u00a0\u2013\u00a0a3\u00a0= (-7.2)-(-5.2) = -2
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or common difference is same every time.
\nTherefore,\u00a0d\u00a0= -2 and the given series are in A.P.
\nHence, next three terms are;
\na5\u00a0= \u2013 7.2-2 = -9.2
\na6\u00a0= \u2013 9.2-2 = \u2013 11.2
\na7\u00a0= \u2013 11.2-2 = \u2013 13.2<\/h3>\n

(iv) Given, -10, \u2013 6, \u2013 2, 2 \u2026<\/h3>\n

Here, the terms and their difference are;
\na2\u00a0\u2013\u00a0a1\u00a0= (-6)-(-10) = 4
\na3\u00a0\u2013\u00a0a2\u00a0= (-2)-(-6) = 4
\na4\u00a0\u2013\u00a0a3\u00a0= (2 -(-2) = 4
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0= 4 and the given numbers are in A.P.
\nHence, next three terms are;
\na5\u00a0= 2+4 = 6
\na6\u00a0= 6+4 = 10
\na7\u00a0= 10+4 = 14<\/h3>\n

(v) Given, 3, 3+\u221a2, 3+2\u221a2, 3+3\u221a2<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0= 3+\u221a2-3 = \u221a2
\na3\u00a0\u2013\u00a0a2\u00a0= (3+2\u221a2)-(3+\u221a2) = \u221a2
\na4\u00a0\u2013\u00a0a3\u00a0= (3+3\u221a2) \u2013 (3+2\u221a2) = \u221a2
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0=\u00a0\u221a2\u00a0and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0= (3+\u221a2)\u00a0+\u221a2\u00a0= 3+4\u221a2
\na6\u00a0= (3+4\u221a2)+\u221a2\u00a0= 3+5\u221a2
\na7\u00a0= (3+5\u221a2)+\u221a2\u00a0= 3+6\u221a2
\n(vi)\u00a00.2, 0.22, 0.222, 0.2222 \u2026.<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a00.22-0.2 = 0.02
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a00.222-0.22 = 0.002
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a00.2222-0.222 = 0.0002
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not same every time.
\nTherefore, and the given series doesn\u2019t forms a A.P.<\/h3>\n

(vii)\u00a00, -4, -8, -12 \u2026<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a0(-4)-0 = -4
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a0(-8)-(-4) = -4
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a0(-12)-(-8) = -4
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0=\u00a0-4\u00a0and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0=\u00a0-12-4 = -16
\na6\u00a0=\u00a0-16-4 = -20
\na7\u00a0=\u00a0-20-4 = -24<\/h3>\n

(viii) -1\/2, -1\/2, -1\/2, -1\/2 \u2026.<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0= (-1\/2) \u2013 (-1\/2) = 0
\na3\u00a0\u2013\u00a0a2\u00a0= (-1\/2) \u2013 (-1\/2) = 0
\na4\u00a0\u2013\u00a0a3\u00a0= (-1\/2) \u2013 (-1\/2) = 0
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0= 0 and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0= (-1\/2)-0 = -1\/2
\na6\u00a0=\u00a0(-1\/2)-0 = -1\/2
\na7\u00a0=\u00a0(-1\/2)-0 = -1\/2<\/h3>\n

(ix) 1, 3, 9, 27 \u2026<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a03-1 = 2
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a09-3 = 6
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a027-9 = 18
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not same every time.
\nTherefore, and the given series doesn\u2019t form a A.P.<\/h3>\n

(x)\u00a0a, 2a, 3a, 4a\u00a0\u2026
\nHere,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a02a\u2013a\u00a0=\u00a0a
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a03a-2a\u00a0=\u00a0a
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a04a-3a\u00a0=\u00a0a
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0=\u00a0a\u00a0and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0=\u00a04a+a\u00a0= 5a
\na6\u00a0= 5a+a\u00a0= 6a
\na7\u00a0=\u00a06a+a\u00a0= 7a<\/h3>\n

(xi)\u00a0a,\u00a0a2,\u00a0a3,\u00a0a4\u00a0\u2026
\nHere,
\na2\u00a0\u2013\u00a0a1\u00a0=\u00a0a2\u2013a\u00a0= a(a-1)
\na3\u00a0\u2013\u00a0a2\u00a0=\u00a0a3\u00a0\u2013\u00a0a2\u00a0=\u00a0a2(a-1)
\na4\u00a0\u2013\u00a0a3\u00a0=\u00a0a4\u00a0\u2013\u00a0a3\u00a0=\u00a0a3(a-1)
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not same every time.
\nTherefore, the given series doesn\u2019t forms a A.P.<\/h3>\n

(xii) \u221a2, \u221a8, \u221a18, \u221a32\u00a0\u2026<\/h3>\n

Here,
\na2\u00a0\u2013\u00a0a1\u00a0= \u221a8-\u221a2\u00a0\u00a0= 2\u221a2-\u221a2\u00a0= \u221a2
\na3\u00a0\u2013\u00a0a2\u00a0= \u221a18-\u221a8\u00a0= 3\u221a2-2\u221a2\u00a0= \u221a2
\na4\u00a0\u2013\u00a0a3\u00a0= 4\u221a2-3\u221a2\u00a0= \u221a2
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0=\u00a0\u221a2\u00a0and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0= \u221a32+\u221a2\u00a0= 4\u221a2+\u221a2\u00a0= 5\u221a2\u00a0= \u221a50
\na6\u00a0= 5\u221a2+\u221a2\u00a0= 6\u221a2\u00a0= \u221a72
\na7\u00a0= 6\u221a2+\u221a2\u00a0= 7\u221a2\u00a0= \u221a98<\/h3>\n

(xiii)\u00a0\u221a3, \u221a6, \u221a9, \u221a12\u00a0\u2026
\nHere,
\na2\u00a0\u2013\u00a0a1\u00a0= \u221a6-\u221a3\u00a0= \u221a3\u00d7\u221a2-\u221a3\u00a0= \u221a3(\u221a2-1)
\na3\u00a0\u2013\u00a0a2\u00a0= \u221a9-\u221a6\u00a0= 3-\u221a6\u00a0= \u221a3(\u221a3-\u221a2)
\na4\u00a0\u2013\u00a0a3\u00a0= \u221a12\u00a0\u2013 \u221a9\u00a0= 2\u221a3\u00a0\u2013 \u221a3\u00d7\u221a3 = \u221a3(2-\u221a3)
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not same every time.
\nTherefore, the given series doesn\u2019t form a A.P.<\/h3>\n

(xiv) 12, 32, 52, 72\u00a0\u2026
\nOr, 1, 9, 25, 49 \u2026..
\nHere,
\na2\u00a0\u2212\u00a0a1\u00a0= 9\u22121 = 8
\na3\u00a0\u2212\u00a0a2\u00a0= 25\u22129 = 16
\na4\u00a0\u2212\u00a0a3\u00a0= 49\u221225 = 24
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is not same every time.
\nTherefore, the given series doesn\u2019t form a A.P.<\/h3>\n

(xv) 12, 52, 72, 73 \u2026
\nOr 1, 25, 49, 73 \u2026
\nHere,
\na2\u00a0\u2212\u00a0a1\u00a0= 25\u22121 = 24
\na3\u00a0\u2212\u00a0a2\u00a0= 49\u221225 = 24
\na4\u00a0\u2212\u00a0a3\u00a0= 73\u221249 = 24
\nSince,\u00a0an+1\u00a0\u2013\u00a0an\u00a0or the common difference is same every time.
\nTherefore,\u00a0d\u00a0= 24\u00a0and the given series forms a A.P.
\nHence, next three terms are;
\na5\u00a0= 73+24 = 97
\na6\u00a0= 97+24 = 121
\na7\u00a0= 121+24 = 145<\/h3>\n

Exercise 5.2<\/h2>\n

1. Fill in the blanks in the following table, given that\u00a0a\u00a0is the first term,\u00a0d\u00a0the common difference and\u00a0an\u00a0the\u00a0nth\u00a0term of the A.P.<\/h2>\n

\"\"
\nSolutions:
\n(i)\u00a0Given,
\nFirst term,\u00a0a\u00a0= 7
\nCommon difference,\u00a0d\u00a0= 3
\nNumber of terms,\u00a0n\u00a0= 8,
\nWe have to find the nth term,\u00a0an\u00a0= ?
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\nPutting the values,
\n=> 7+(8 \u22121) 3
\n=> 7+(7) 3
\n=> 7+21 = 28
\nHence,\u00a0an\u00a0= 28<\/h3>\n

(ii)\u00a0Given,
\nFirst term,\u00a0a\u00a0= -18
\nCommon difference,\u00a0d\u00a0= ?
\nNumber of terms,\u00a0n\u00a0= 10
\nNth term,\u00a0an\u00a0= 0
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\nPutting the values,
\n0 = \u2212 18 +(10\u22121)d
\n18 = 9d
\nd\u00a0= 18\/9 = 2
\nHence, common difference,\u00a0d\u00a0= 2<\/h3>\n

(iii)\u00a0Given,
\nFirst term,\u00a0a\u00a0= ?
\nCommon difference,\u00a0d\u00a0= -3
\nNumber of terms,\u00a0n\u00a0= 18
\nNth term,\u00a0an\u00a0= -5
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\nPutting the values,
\n\u22125 =\u00a0a+(18\u22121) (\u22123)
\n\u22125 =\u00a0a+(17) (\u22123)
\n\u22125 =\u00a0a\u221251
\na\u00a0= 51\u22125 = 46
\nHence,\u00a0a\u00a0= 46<\/h3>\n

(iv)\u00a0Given,
\nFirst term,\u00a0a\u00a0= -18.9
\nCommon difference,\u00a0d\u00a0= 2.5
\nNumber of terms,\u00a0n\u00a0= ?
\nNth term,\u00a0an\u00a0= 3.6
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a\u00a0+(n\u00a0\u22121)d
\nPutting the values,
\n3.6 = \u2212 18.9+(n\u00a0\u22121)2.5
\n3.6 + 18.9 = (n\u22121)2.5
\n22.5 = (n\u22121)2.5
\n(n\u00a0\u2013 1) = 22.5\/2.5
\nn\u00a0\u2013 1 = 9
\nn\u00a0= 10
\nHence,\u00a0n\u00a0= 10<\/h3>\n

(v)\u00a0Given,
\nFirst term,\u00a0a\u00a0= 3.5
\nCommon difference,\u00a0d\u00a0= 0
\nNumber of terms,\u00a0n\u00a0= 105
\nNth term,\u00a0an\u00a0= ?
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d
\nPutting the values,
\nan\u00a0= 3.5+(105\u22121) 0
\nan\u00a0= 3.5+104\u00d70
\nan\u00a0= 3.5
\nHence,\u00a0an\u00a0= 3.5<\/h3>\n

2. Choose the correct choice in the following and justify:
\n(i) 30th\u00a0term of the A.P: 10,7, 4, \u2026, is
\n(A)\u00a097 (B)\u00a077 (C)\u00a0\u221277 (D) \u221287
\n(ii) 11th\u00a0term of the A.P. -3, -1\/2, ,2 \u2026. is
\n(A) 28 (B) 22 (C) \u2013 38 (D)<\/h3>\n

\"\"<\/p>\n

Solutions:
\n(i) Given here,
\nA.P. = 10, 7, 4, \u2026
\nTherefore, we can find,
\nFirst term,\u00a0a\u00a0= 10
\nCommon difference,\u00a0d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 7\u221210 = \u22123
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a\u00a0+(n\u22121)d
\nPutting the values;
\na30\u00a0= 10+(30\u22121)(\u22123)
\na30\u00a0= 10+(29)(\u22123)
\na30\u00a0= 10\u221287 = \u221277
\nHence, the correct answer is option\u00a0C.<\/h3>\n

(ii) Given here,
\nA.P. =\u00a0-3, -1\/2, ,2 \u2026
\nTherefore, we can find,
\nFirst term\u00a0a\u00a0= \u2013 3
\nCommon difference,\u00a0d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= (-1\/2) -(-3)
\n\u21d2(-1\/2)\u00a0+ 3 = 5\/2
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\nPutting the values;
\na11\u00a0= -3+(11-1)(5\/2)
\na11\u00a0= -3+(10)(5\/2)
\na11\u00a0= -3+25
\na11\u00a0= 22
\nHence, the answer is option B.<\/h3>\n

3. In the following APs find the missing term in the boxes.<\/h2>\n

\"\"
\nSolutions:
\n(i)\u00a0For the given A.P., 2,2 , 26
\nThe first and third term are;
\na\u00a0= 2
\na3\u00a0= 26
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d
\nTherefore, putting the values here,
\na3\u00a0= 2+(3-1)d
\n26 = 2+2d
\n24 = 2d
\nd\u00a0= 12
\na2\u00a0= 2+(2-1)12
\n= 14
\nTherefore, 14 is the missing term.<\/h3>\n

(ii)\u00a0For the given A.P., , 13, ,3
\na2\u00a0= 13 and
\na4\u00a0= 3
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)\u00a0d
\nTherefore, putting the values here,
\na2\u00a0=\u00a0a\u00a0+(2-1)d
\n13 =\u00a0a+d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\na4\u00a0=\u00a0a+(4-1)d
\n3 =\u00a0a+3d\u00a0\u2026\u2026\u2026\u2026..\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get,
\n\u2013 10 = 2d
\nd\u00a0= \u2013 5
\nFrom equation\u00a0(i), putting the value of d,we get
\n13 =\u00a0a+(-5)
\na\u00a0= 18
\na3\u00a0= 18+(3-1)(-5)
\n= 18+2(-5) = 18-10 = 8
\nTherefore, the missing terms are 18 and 8 respectively.<\/h3>\n

(iii)\u00a0For the given A.P.,
\na\u00a0= 5 and
\na4\u00a0= 19\/2
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore, putting the values here,
\na4\u00a0=\u00a0a+(4-1)d
\n19\/2 =\u00a05+3d
\n(19\/2) \u2013 5 = 3d
\n3d =\u00a09\/2
\nd = 3\/2
\na2\u00a0=\u00a0a+(2-1)d
\na2\u00a0=\u00a05+3\/2
\na2\u00a0=\u00a013\/2
\na3\u00a0=\u00a0a+(3-1)d
\na3\u00a0=\u00a05+2\u00d73\/2
\na3\u00a0=\u00a08
\nTherefore, the missing terms are 13\/2 and 8 respectively.<\/h3>\n

(iv) For the given A.P.,
\na\u00a0= \u22124 and
\na6\u00a0= 6
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a\u00a0+(n\u22121)\u00a0d
\nTherefore, putting the values here,
\na6\u00a0= a+(6\u22121)d
\n6 = \u2212 4+5d
\n10 = 5d
\nd\u00a0= 2
\na2\u00a0=\u00a0a+d\u00a0= \u2212 4+2 = \u22122
\na3\u00a0=\u00a0a+2d\u00a0= \u2212 4+2(2) = 0
\na4\u00a0=\u00a0a+3d\u00a0= \u2212 4+ 3(2) = 2
\na5\u00a0=\u00a0a+4d\u00a0= \u2212 4+4(2) = 4
\nTherefore, the missing terms are \u22122, 0, 2, and 4 respectively.<\/h3>\n

(v)\u00a0For the given A.P.,
\na2\u00a0= 38
\na6\u00a0= \u221222
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d
\nTherefore, putting the values here,
\na2\u00a0=\u00a0a+(2\u22121)d
\n38 =\u00a0a+d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\na6\u00a0=\u00a0a+(6\u22121)d
\n\u221222 =\u00a0a+5d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get
\n\u2212 22 \u2212 38 = 4d
\n\u221260 = 4d
\nd\u00a0= \u221215
\na\u00a0=\u00a0a2\u00a0\u2212\u00a0d\u00a0= 38 \u2212 (\u221215) = 53
\na3\u00a0=\u00a0a\u00a0+ 2d\u00a0= 53 + 2 (\u221215) = 23
\na4\u00a0=\u00a0a\u00a0+ 3d\u00a0= 53 + 3 (\u221215) = 8
\na5\u00a0=\u00a0a\u00a0+ 4d\u00a0= 53 + 4 (\u221215) = \u22127
\nTherefore, the missing terms are 53, 23, 8, and \u22127 respectively.<\/h3>\n

4. Which term of the A.P. 3, 8, 13, 18, \u2026 is 78?<\/h2>\n

Solutions:
\nGiven the A.P. series as3, 8, 13, 18, \u2026
\nFirst term, a\u00a0= 3
\nCommon difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 8 \u2212 3 = 5
\nLet\u00a0the\u00a0nth\u00a0term of given A.P. be 78. Now as we know,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\n78 = 3+(n\u00a0\u22121)5
\n75 = (n\u22121)5
\n(n\u22121) = 15
\nn\u00a0= 16
\nHence, 16th\u00a0term of this A.P. is 78.<\/h3>\n

5. Find the number of terms in each of the following A.P.
\n(i) 7, 13, 19, \u2026, 205<\/h2>\n

\"\"
\nSolutions:
\n(i) Given, 7, 13, 19, \u2026, 205 is the A.P
\nTherefore
\nFirst term, a\u00a0= 7
\nCommon difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 13 \u2212 7 = 6
\nLet there are\u00a0n\u00a0terms in this A.P.
\nan\u00a0= 205
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a\u00a0+ (n\u00a0\u2212 1)\u00a0d
\nTherefore, 205 = 7 + (n\u00a0\u2212 1) 6
\n198 = (n\u00a0\u2212 1) 6
\n33 = (n\u00a0\u2212 1)
\nn\u00a0= 34
\nTherefore, this given series has 34 terms in it.<\/h3>\n

\"\"
\nFirst term, a\u00a0= 18
\nCommon difference, d = a2-a1\u00a0=<\/h3>\n

\"\"
\nd = (31-36)\/2 = -5\/2
\nLet there are n terms in this A.P.
\nan\u00a0= -47
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\n-47 = 18+(n-1)(-5\/2)
\n-47-18 = (n-1)(-5\/2)
\n-65 = (n-1)(-5\/2)
\n(n-1) = -130\/-5
\n(n-1) =\u00a026
\nn\u00a0= 27
\nTherefore, this given A.P. has 27 terms in it.<\/h3>\n

6.\u00a0Check whether -150 is a term of the A.P. 11, 8, 5, 2, \u2026<\/h2>\n

Solution:
\nFor the given series, A.P. 11, 8, 5, 2..
\nFirst term, a\u00a0= 11
\nCommon difference, d\u00a0=\u00a0a2\u2212a1\u00a0= 8\u221211 = \u22123
\nLet \u2212150 be the\u00a0nth\u00a0term of this A.P.
\nAs we know, for an A.P.,
\nan\u00a0=\u00a0a+(n\u22121)d
\n-150 = 11+(n\u00a0-1)(-3)
\n-150 = 11-3n\u00a0+3
\n-164 = -3n
\nn\u00a0= 164\/3
\nClearly,\u00a0n\u00a0is not an integer but a fraction.
\nTherefore, \u2013 150 is not a term of this A.P.<\/h3>\n

7. Find the 31st\u00a0term of an A.P. whose 11th\u00a0term is 38 and the 16th\u00a0term is 73.<\/h2>\n

Solution:
\nGiven that,
\n11th\u00a0term, a11\u00a0= 38
\nand 16th\u00a0term, a16\u00a0= 73
\nWe know that,
\nan\u00a0=\u00a0a+(n\u22121)d
\na11\u00a0=\u00a0a+(11\u22121)d
\n38 =\u00a0a+10d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nIn the same way,
\na16\u00a0=\u00a0a\u00a0+(16\u22121)d
\n73 =\u00a0a+15d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get
\n35 = 5d
\nd\u00a0= 7
\nFrom equation\u00a0(i), we can write,
\n38 =\u00a0a+10\u00d7(7)
\n38 \u2212 70 =\u00a0a
\na\u00a0= \u221232
\na31\u00a0=\u00a0a\u00a0+(31\u22121)\u00a0d
\n= \u2212 32 + 30 (7)
\n= \u2212 32 + 210
\n= 178
\nHence, 31st\u00a0term is 178.<\/h3>\n

8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.<\/h2>\n

Solution:\u00a0Given that,
\n3rd\u00a0term, a3\u00a0= 12
\n50th\u00a0term, a50\u00a0= 106
\nWe know that,
\nan\u00a0=\u00a0a+(n\u22121)d
\na3\u00a0=\u00a0a+(3\u22121)d
\n12 =\u00a0a+2d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nIn the same way,
\na50\u00a0=\u00a0a+(50\u22121)d
\n106 =\u00a0a+49d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get
\n94 = 47d
\nd\u00a0= 2 = common difference
\nFrom equation\u00a0(i), we can write now,
\n12 =\u00a0a+2(2)
\na\u00a0= 12\u22124 = 8
\na29\u00a0=\u00a0a+(29\u22121)\u00a0d
\na29\u00a0= 8+(28)2
\na29\u00a0= 8+56 = 64
\nTherefore, 29th\u00a0term is 64.<\/h3>\n

9. If the 3rd\u00a0and the 9th\u00a0terms of an A.P. are 4 and \u2212 8 respectively. Which term of this A.P. is zero.<\/h2>\n

Solution:
\nGiven that,
\n3rd\u00a0term, a3\u00a0= 4
\nand 9th\u00a0term, a9\u00a0= \u22128
\nWe know that,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\na3\u00a0=\u00a0a+(3\u22121)d
\n4 =\u00a0a+2d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u00a0(i)
\na9\u00a0=\u00a0a+(9\u22121)d
\n\u22128 =\u00a0a+8d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we will get here,
\n\u221212 = 6d
\nd\u00a0= \u22122
\nFrom equation\u00a0(i), we can write,
\n4 =\u00a0a+2(\u22122)
\n4 =\u00a0a\u22124
\na\u00a0= 8
\nLet\u00a0nth\u00a0term of this A.P. be zero.
\nan\u00a0=\u00a0a+(n\u22121)d
\n0 = 8+(n\u22121)(\u22122)
\n0 = 8\u22122n+2
\n2n\u00a0= 10
\nn\u00a0= 5
\nHence, 5th\u00a0term of this A.P. is 0.<\/h3>\n

10. If 17th\u00a0term of an A.P. exceeds its 10th\u00a0term by 7. Find the common difference.<\/h2>\n

Solution:
\nWe know that, for an A.P series;
\nan\u00a0=\u00a0a+(n\u22121)d
\na17\u00a0=\u00a0a+(17\u22121)d
\na17\u00a0=\u00a0a\u00a0+16d
\nIn the same way,
\na10\u00a0=\u00a0a+9d
\nAs it is given in the question,
\na17\u00a0\u2212\u00a0a10\u00a0= 7
\nTherefore,
\n(a\u00a0+16d)\u2212(a+9d) = 7
\n7d\u00a0= 7
\nd\u00a0= 1
\nTherefore, the common difference is 1.<\/h3>\n

11.\u00a0Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th\u00a0term?<\/h2>\n

Solution:
\nGiven A.P. is 3, 15, 27, 39, \u2026
\nfirst term, a\u00a0= 3
\ncommon difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 15 \u2212 3 = 12
\nWe know that,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\na54\u00a0=\u00a0a+(54\u22121)d
\n\u21d23+(53)(12)
\n\u21d23+636 = 639
\na54\u00a0= 639+132=771
\nWe have to find the term of this A.P. which is 132 more than a54,\u00a0i.e.771.
\nLet\u00a0nth\u00a0term be 771.
\nan\u00a0=\u00a0a+(n\u22121)d
\n771 = 3+(n\u00a0\u22121)12
\n768 = (n\u22121)12
\n(n\u00a0\u22121) = 64
\nn\u00a0= 65
\nTherefore, 65th\u00a0term was 132 more than 54th\u00a0term.
\nOr another method is;<\/h3>\n

Let\u00a0nth\u00a0term be 132 more than 54th\u00a0term.
\nn\u00a0= 54\u00a0+ 132\/2
\n= 54\u00a0+ 11 =\u00a065th\u00a0term<\/h3>\n

12. Two APs have the same common difference. The difference between their 100th\u00a0term is 100, what is the difference between their 1000th\u00a0terms?<\/h2>\n

Solution:
\nLet, the first term of two APs be\u00a0a1\u00a0and\u00a0a2\u00a0respectively
\nAnd the common difference of these APs be\u00a0d.
\nFor the first A.P.,we know,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\na100\u00a0=\u00a0a1+(100\u22121)d
\n=\u00a0a1\u00a0+ 99d
\na1000\u00a0=\u00a0a1+(1000\u22121)d
\na1000\u00a0=\u00a0a1+999d
\nFor second A.P., we know,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\na100\u00a0=\u00a0a2+(100\u22121)d
\n=\u00a0a2+99d
\na1000\u00a0=\u00a0a2+(1000\u22121)d
\n=\u00a0a2+999d
\nGiven that, difference between 100th\u00a0term of the two APs = 100
\nTherefore, (a1+99d) \u2212 (a2+99d) = 100
\na1\u2212a2\u00a0= 100\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(i)
\nDifference between 1000th\u00a0terms of the two APs
\n(a1+999d) \u2212 (a2+999d) =\u00a0a1\u2212a2
\nFrom equation\u00a0(i),
\nThis difference,\u00a0a1\u2212a2\u00a0= 100
\nHence, the difference between 1000th\u00a0terms of the two A.P. will be 100.<\/h3>\n

13. How many three digit numbers are divisible by 7?<\/h2>\n

Solution:
\nFirst three-digit number that is divisible by 7 are;
\nFirst number = 105
\nSecond number = 105+7 = 112
\nThird number = 112+7 =119
\nTherefore, 105, 112, 119, \u2026
\nAll are three digit numbers are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
\nAs we know, the largest possible three-digit number is 999.
\nWhen we divide 999 by 7, the remainder will be 5.
\nTherefore, 999-5 = 994 is the maximum possible three-digit number that is divisible by 7.
\nNow the series is as follows.
\n105, 112, 119, \u2026, 994
\nLet 994 be the\u00a0nth term of this A.P.
\nfirst term, a\u00a0= 105
\ncommon difference, d\u00a0= 7
\nan\u00a0= 994
\nn\u00a0= ?
\nAs we know,
\nan\u00a0=\u00a0a+(n\u22121)d
\n994 = 105+(n\u22121)7
\n889 = (n\u22121)7
\n(n\u22121) = 127
\nn\u00a0= 128
\nTherefore, 128 three-digit numbers are divisible by 7.<\/h3>\n

14. How many multiples of 4 lie between 10 and 250?<\/h2>\n

Solution:
\nThe first multiple of 4 that is greater than 10 is 12.
\nNext multiple will be 16.
\nTherefore, the series formed as;
\n12, 16, 20, 24, \u2026
\nAll these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
\nWhen we divide 250 by 4, the remainder will be 2. Therefore, 250 \u2212 2 = 248 is divisible by 4.
\nThe series is as follows, now;
\n12, 16, 20, 24, \u2026, 248
\nLet 248 be the\u00a0nth\u00a0term of this A.P.
\nfirst term, a\u00a0= 12
\ncommon difference, d\u00a0= 4
\nan\u00a0=\u00a0248
\nAs we know,
\nan\u00a0=\u00a0a+(n\u22121)d
\n248 = 12+(n-1)\u00d74
\n236\/4 =\u00a0n-1
\n59 \u00a0=\u00a0n-1
\nn\u00a0= 60
\nTherefore, there are 60 multiples of 4 between 10 and 250.<\/h3>\n

15. For what value of\u00a0n, are the\u00a0nth\u00a0terms of two APs 63, 65, 67, and 3, 10, 17, \u2026 equal?<\/h2>\n

Solution:
\nGiven two APs as; 63, 65, 67,\u2026 and 3, 10, 17,\u2026.
\nTaking first AP,
\n63, 65, 67, \u2026
\nFirst term, a\u00a0= 63
\nCommon difference, d\u00a0=\u00a0a2\u2212a1\u00a0= 65\u221263 = 2
\nWe know, nth\u00a0term of this A.P. =\u00a0an\u00a0=\u00a0a+(n\u22121)d
\nan= 63+(n\u22121)2 = 63+2n\u22122
\nan\u00a0= 61+2n\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nTaking second AP,
\n3, 10, 17, \u2026
\nFirst term, a\u00a0= 3
\nCommon difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 10 \u2212 3 = 7
\nWe know that,
\nnth\u00a0term of this A.P. = 3+(n\u22121)7
\nan\u00a0= 3+7n\u22127
\nan\u00a0= 7n\u22124 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(ii)
\nGiven,\u00a0nth\u00a0term of these A.P.s are equal to each other.
\nEquating both these equations, we get,
\n61+2n\u00a0= 7n\u22124
\n61+4 = 5n
\n5n\u00a0= 65
\nn\u00a0= 13
\nTherefore, 13th\u00a0terms of both these A.P.s are equal to each other.<\/h3>\n

 <\/p>\n

16. Determine the A.P. whose third term is 16 and the 7th\u00a0term exceeds the 5th\u00a0term by 12.<\/h2>\n

Solutions:
\nGiven,
\nThird term, a3\u00a0= 16
\nAs we know,
\na\u00a0+(3\u22121)d\u00a0= 16
\na+2d\u00a0= 16 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nIt is given that, 7th\u00a0term exceeds the 5th\u00a0term by 12.
\na7\u00a0\u2212\u00a0a5\u00a0= 12
\n[a+(7\u22121)d]\u2212[a\u00a0+(5\u22121)d]= 12
\n(a+6d)\u2212(a+4d) = 12
\n2d\u00a0= 12
\nd\u00a0= 6
\nFrom equation\u00a0(i), we get,
\na+2(6) = 16
\na+12 = 16
\na\u00a0= 4
\nTherefore, A.P. will be4, 10, 16, 22, \u2026<\/h3>\n

17. Find the 20th\u00a0term from the last term of the A.P. 3, 8, 13, \u2026, 253.<\/h2>\n

Solution:
\nGiven A.P. is3, 8, 13, \u2026, 253
\nCommon difference, d= 5.
\nTherefore, we can write the given AP in reverse order as;
\n253, 248, 243, \u2026, 13, 8, 5
\nNow for the new AP,
\nfirst term, a\u00a0= 253
\nand common difference, d\u00a0= 248 \u2212 253 = \u22125
\nn\u00a0= 20
\nTherefore, using nth term formula, we get,
\na20\u00a0=\u00a0a+(20\u22121)d
\na20\u00a0= 253+(19)(\u22125)
\na20\u00a0= 253\u221295
\na\u00a0= 158
\nTherefore, 20th\u00a0term from the last term of the AP 3, 8, 13, \u2026, 253.is 158.<\/h3>\n

18. The sum of 4th\u00a0and 8th\u00a0terms of an A.P. is 24 and the sum of the 6th\u00a0and 10th\u00a0terms is 44. Find the first three terms of the A.P.<\/h2>\n

Solution:
\nWe know that, the nth term of the AP is;
\nan\u00a0=\u00a0a+(n\u22121)d
\na4\u00a0=\u00a0a+(4\u22121)d
\na4\u00a0=\u00a0a+3d
\nIn the same way, we can write,
\na8\u00a0=\u00a0a+7d
\na6\u00a0=\u00a0a+5d
\na10\u00a0=\u00a0a+9d
\nGiven that,
\na4+a8\u00a0= 24
\na+3d+a+7d\u00a0= 24
\n2a+10d\u00a0= 24
\na+5d\u00a0= 12 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u00a0(i)
\na6+a10\u00a0= 44
\na\u00a0+5d+a+9d\u00a0= 44
\n2a+14d\u00a0= 44
\na+7d\u00a0= 22 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(ii)
\nOn subtracting equation\u00a0(i)\u00a0from\u00a0(ii), we get,
\n2d\u00a0= 22 \u2212 12
\n2d\u00a0= 10
\nd\u00a0= 5
\nFrom equation\u00a0(i), we get,
\na+5d\u00a0= 12
\na+5(5) = 12
\na+25 = 12
\na\u00a0= \u221213
\na2\u00a0=\u00a0a+d\u00a0= \u2212 13+5 = \u22128
\na3\u00a0=\u00a0a2+d\u00a0= \u2212 8+5 = \u22123
\nTherefore, the first three terms of this A.P. are \u221213, \u22128, and \u22123.<\/h3>\n

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?<\/h2>\n

Solution:
\nIt can be seen from the given question, that the incomes of Subba Rao increases every year by Rs.200 and hence, forms an AP.
\nTherefore, after 1995, the salaries of each year are;
\n5000, 5200, 5400, \u2026
\nHere,\u00a0first term,\u00a0a\u00a0= 5000
\nand common difference, d\u00a0= 200
\nLet after\u00a0nth\u00a0year, his salary be Rs 7000.
\nTherefore, by the nth\u00a0term formula of AP,
\nan\u00a0=\u00a0a+(n\u22121)\u00a0d
\n7000 = 5000+(n\u22121)200
\n200(n\u22121)= 2000
\n(n\u22121) = 10
\nn\u00a0= 11
\nTherefore, in 11th year, his salary will be Rs 7000.<\/h3>\n

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the\u00a0nth\u00a0week, her weekly savings become Rs 20.75, find\u00a0n.<\/h2>\n

Solution:
\nGiven that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75.
\nHence,
\nFirst term, a\u00a0= 5
\nand common difference, d\u00a0= 1.75
\nAlso given,
\nan\u00a0= 20.75
\nFind, n\u00a0= ?
\nAs we know, by the nth\u00a0term formula,
\nan\u00a0=\u00a0a+(n\u22121)d
\nTherefore,
\n20.75 = 5+(n\u00a0-1)\u00d71.75
\n15.75 = (n\u00a0-1)\u00d71.75
\n(n\u00a0-1) = 15.75\/1.75 = 1575\/175
\n= 63\/7 = 9
\nn\u00a0-1 = 9
\nn\u00a0= 10
\nHence,\u00a0n\u00a0is 10.<\/h3>\n

Exercise 5.3<\/h2>\n

1. Find the sum of the following APs.
\n(i) 2, 7, 12 ,\u2026., to 10 terms.
\n(ii) \u2212 37, \u2212 33, \u2212 29 ,\u2026, to 12 terms
\n(iii) 0.6, 1.7, 2.8 ,\u2026\u2026.., to 100 terms
\n(iv) 1\/15, 1\/12, 1\/10, \u2026\u2026 , to 11 terms<\/h2>\n

Solutions:
\n(i)\u00a0Given, 2, 7, 12 ,\u2026, to 10 terms
\nFor this A.P.,
\nfirst term, a\u00a0= 2
\nAnd common difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 7\u22122 = 5
\nn\u00a0= 10
\nWe know that, the formula for sum of nth term in AP series is,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n-1)d]
\nS10\u00a0= 10\/2\u00a0[2(2)+(10\u00a0-1)\u00d75]
\n= 5[4+(9)\u00d7(5)]
\n= 5 \u00d7 49 = 245<\/h3>\n

(ii)\u00a0Given, \u221237, \u221233, \u221229 ,\u2026, to 12 terms
\nFor this A.P.,
\nfirst term, a\u00a0= \u221237
\nAnd common difference, d\u00a0=\u00a0a2\u2212\u00a0a1
\nd= (\u221233)\u2212(\u221237)
\n= \u2212 33 + 37 = 4
\nn\u00a0= 12
\nWe know that, the formula for sum of nth term in AP series is,
\nSn\u00a0=\u00a0n\/2\u00a0[2a+(n-1)d]
\nS12\u00a0= 12\/2\u00a0[2(-37)+(12-1)\u00d74]
\n= 6[-74+11\u00d74]
\n= 6[-74+44]
\n= 6(-30) = -180<\/h3>\n

(iii)\u00a0Given, 0.6, 1.7, 2.8 ,\u2026, to 100 terms
\nFor this A.P.,
\nfirst term, a\u00a0= 0.6
\nCommon difference, d\u00a0=\u00a0a2\u00a0\u2212\u00a0a1\u00a0= 1.7 \u2212 0.6 = 1.1
\nn\u00a0= 100
\nWe know that, the formula for sum of nth term in AP series is,
\nSn\u00a0=\u00a0n\/2[2a\u00a0+(n-1)d]
\nS12\u00a0= 50\/2\u00a0[1.2+(99)\u00d71.1]
\n= 50[1.2+108.9]
\n= 50[110.1]
\n= 5505<\/h3>\n

(iv) Given, 1\/15, 1\/12, 1\/10, \u2026\u2026 , to 11 terms
\nFor this A.P.,
\nFirst term, a = 1\/5
\nCommon difference, d = a2\u00a0\u2013a1\u00a0= (1\/12)-(1\/5) = 1\/60
\nAnd number of terms n = 11
\nWe know that, the formula for sum of nth term in AP series is,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+\u00a0(n\u00a0\u2013 1)\u00a0d]<\/h3>\n

\"\"
\n= 11\/2(2\/15 + 10\/60)
\n= 11\/2 (9\/30)
\n= 33\/20<\/h3>\n

2. Find the sums given below:<\/h2>\n

\"\"
\n(ii) 34 + 32 + 30 + \u2026\u2026\u2026.. + 10
\n(iii) \u2212 5 + (\u2212 8) + (\u2212 11) + \u2026\u2026\u2026\u2026 + (\u2212 230)<\/h2>\n

Solutions:
\n(i)<\/h3>\n

\"\"
\nFirst term, a\u00a0= 7
\nnth\u00a0term, an\u00a0= 84<\/h3>\n

\"\"
\nLet 84 be the\u00a0nth\u00a0term of this A.P., then as per the nth\u00a0term formula,
\nan\u00a0=\u00a0a(n-1)d
\n84 = 7+(n\u00a0\u2013 1)\u00d77\/2
\n77 = (n-1)\u00d77\/2
\n22 =\u00a0n\u22121
\nn\u00a0= 23
\nWe know that, sum of n term is;
\nSn\u00a0=\u00a0n\/2 (a\u00a0+\u00a0l) , l = 84
\nSn\u00a0= 23\/2\u00a0(7+84)
\nSn\u00a0 = (23\u00d791\/2) = 2093\/2<\/h3>\n

\"\"<\/h3>\n

(ii)\u00a0Given, 34 + 32 + 30 + \u2026\u2026\u2026.. + 10
\nFor this A.P.,
\nfirst term, a\u00a0= 34
\ncommon difference, d\u00a0=\u00a0a2\u2212a1\u00a0= 32\u221234 = \u22122
\nnth\u00a0term, an= 10
\nLet 10 be the\u00a0nth\u00a0term of this A.P., therefore,
\nan=\u00a0a\u00a0+(n\u22121)d
\n10 = 34+(n\u22121)(\u22122)
\n\u221224 = (n\u00a0\u22121)(\u22122)
\n12 =\u00a0n\u00a0\u22121
\nn\u00a0= 13
\nWe know that, sum of n terms is;
\nSn\u00a0=\u00a0n\/2 (a\u00a0+l) , l = 10
\n= 13\/2 (34\u00a0+ 10)
\n= (13\u00d744\/2) = 13 \u00d7 22
\n= 286<\/h3>\n

(iii)\u00a0Given, (\u22125) + (\u22128) + (\u221211) + \u2026\u2026\u2026\u2026 + (\u2212230)
\nFor this A.P.,
\nFirst term, a\u00a0= \u22125
\nnth term,\u00a0an= \u2212230
\nCommon difference, d\u00a0=\u00a0a2\u2212a1\u00a0= (\u22128)\u2212(\u22125)
\n\u21d2d = \u2212 8+5 = \u22123
\nLet \u2212230 be the\u00a0nth\u00a0term of this A.P., and by the nth\u00a0term formula we know,
\nan=\u00a0a+(n\u22121)d
\n\u2212230 = \u2212 5+(n\u22121)(\u22123)
\n\u2212225 = (n\u22121)(\u22123)
\n(n\u22121) = 75
\nn\u00a0= 76
\nAnd, Sum of n term,
\nSn\u00a0=\u00a0n\/2 (a\u00a0+\u00a0l)
\n= 76\/2\u00a0[(-5) +\u00a0(-230)]
\n= 38(-235)
\n= -8930<\/h3>\n

3. In an AP
\n(i) Given\u00a0a\u00a0= 5,\u00a0d\u00a0= 3,\u00a0an\u00a0= 50, find\u00a0n\u00a0and\u00a0Sn.
\n(ii) Given\u00a0a\u00a0= 7,\u00a0a13\u00a0= 35, find\u00a0d\u00a0and\u00a0S13.
\n(iii) Given\u00a0a12\u00a0= 37,\u00a0d\u00a0= 3, find\u00a0a\u00a0and\u00a0S12.
\n(iv) Given\u00a0a3\u00a0= 15,\u00a0S10\u00a0= 125, find\u00a0d\u00a0and\u00a0a10.
\n(v) Given\u00a0d\u00a0= 5,\u00a0S9\u00a0= 75, find\u00a0a\u00a0and\u00a0a9.
\n(vi) Given\u00a0a\u00a0= 2,\u00a0d\u00a0= 8,\u00a0Sn\u00a0= 90, find\u00a0n\u00a0and\u00a0an.
\n(vii) Given\u00a0a\u00a0= 8,\u00a0an\u00a0= 62,\u00a0Sn\u00a0= 210, find\u00a0n\u00a0and\u00a0d.
\n(viii) Given\u00a0an\u00a0= 4,\u00a0d\u00a0= 2,\u00a0Sn\u00a0= \u2212 14, find\u00a0n\u00a0and\u00a0a.
\n(ix) Given\u00a0a\u00a0= 3,\u00a0n\u00a0= 8,\u00a0S\u00a0= 192, find\u00a0d.
\n(x) Given\u00a0l\u00a0= 28,\u00a0S\u00a0= 144 and there are total 9 terms. Find\u00a0a.<\/h2>\n

Solutions:
\n(i)\u00a0Given that,\u00a0a\u00a0= 5,\u00a0d\u00a0= 3,\u00a0an\u00a0= 50
\nAs\u00a0we know, from the formula of the nth term in an AP,
\nan\u00a0=\u00a0a\u00a0+(n\u00a0\u22121)d,
\nTherefore, putting the given values, we get,
\n\u21d2 50 = 5+(n\u00a0-1)\u00d73
\n\u21d2 3(n\u00a0-1) = 45
\n\u21d2\u00a0n\u00a0-1 = 15
\n\u21d2\u00a0n\u00a0= 16
\nNow, sum of n terms,
\nSn\u00a0=\u00a0n\/2 (a\u00a0+an)
\nSn\u00a0= 16\/2 (5\u00a0+ 50) = 440<\/h3>\n

(ii)\u00a0Given that,\u00a0a\u00a0= 7,\u00a0a13\u00a0= 35
\nAs\u00a0we know, from the formula of the nth term in an AP,
\nan\u00a0=\u00a0a+(n\u22121)d,
\nTherefore, putting the given values, we get,
\n\u21d2 35 = 7+(13-1)d
\n\u21d2 12d\u00a0= 28
\n\u21d2\u00a0d\u00a0= 28\/12 = 2.33
\nNow,\u00a0Sn\u00a0=\u00a0n\/2 (a+an)
\nS13\u00a0=\u00a013\/2 (7+35) = 273<\/h3>\n

(iii)\u00a0Given that,\u00a0a12\u00a0= 37,\u00a0d\u00a0= 3
\nAs\u00a0we know, from the formula of the nth\u00a0term in an AP,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d,
\nTherefore, putting the given values, we get,
\n\u21d2\u00a0a12\u00a0=\u00a0a+(12\u22121)3
\n\u21d2 37 =\u00a0a+33
\n\u21d2\u00a0a\u00a0= 4
\nNow, sum of nth term,
\nSn\u00a0=\u00a0n\/2 (a+an)
\nSn\u00a0=\u00a012\/2 (4+37)
\n= 246<\/h3>\n

(iv)\u00a0Given that,\u00a0a3\u00a0= 15,\u00a0S10\u00a0= 125
\nAs\u00a0we know, from the formula of the nth term in an AP,
\nan\u00a0=\u00a0a\u00a0+(n\u22121)d,
\nTherefore, putting the given values, we get,
\na3\u00a0=\u00a0a+(3\u22121)d
\n15 =\u00a0a+2d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(i)
\nSum of the nth term,
\nSn\u00a0=\u00a0n\/2\u00a0[2a+(n-1)d]
\nS10\u00a0=\u00a010\/2\u00a0[2a+(10-1)d]
\n125 = 5(2a+9d)
\n25 = 2a+9d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..\u00a0(ii)
\nOn multiplying equation\u00a0(i)\u00a0by\u00a0(ii), we will get;
\n30 = 2a+4d\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(iii)
\nBy subtracting equation\u00a0(iii)\u00a0from\u00a0(ii), we get,
\n\u22125 = 5d
\nd\u00a0= \u22121
\nFrom equation\u00a0(i),
\n15 =\u00a0a+2(\u22121)
\n15 =\u00a0a\u22122
\na\u00a0= 17 = First term
\na10\u00a0=\u00a0a+(10\u22121)d
\na10\u00a0= 17+(9)(\u22121)
\na10\u00a0= 17\u22129 = 8<\/h3>\n

(v)\u00a0Given that,\u00a0d\u00a0= 5,\u00a0S9\u00a0= 75
\nAs, sum of n terms in AP is,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0-1)d]
\nTherefore, the sum of first nine terms are;
\nS9\u00a0= 9\/2\u00a0[2a\u00a0+(9-1)5]
\n25 = 3(a+20)
\n25 = 3a+60
\n3a\u00a0= 25\u221260
\na\u00a0= -35\/3
\nAs we know, the nth\u00a0term can be written as;
\nan\u00a0=\u00a0a+(n\u22121)d
\na9\u00a0=\u00a0a+(9\u22121)(5)
\n= -35\/3+8(5)
\n= -35\/3+40
\n= (35+120\/3) = 85\/3<\/h3>\n

(vi)\u00a0Given that,\u00a0a\u00a0= 2,\u00a0d\u00a0= 8,\u00a0Sn\u00a0= 90
\nAs, sum of n terms in an AP is,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0-1)d]
\n90 =\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0-1)d]
\n\u21d2 180 =\u00a0n(4+8n\u00a0-8) =\u00a0n(8n-4) = 8n2-4n
\n\u21d2 8n2-4n \u2013180 = 0
\n\u21d2 2n2\u2013n-45 = 0
\n\u21d2 2n2-10n+9n-45 = 0
\n\u21d2 2n(n\u00a0-5)+9(n\u00a0-5) = 0
\n\u21d2 (n-5)(2n+9) = 0
\nSo,\u00a0n\u00a0= 5 (as n only be a positive integer)
\n\u2234\u00a0a5\u00a0= 8+5\u00d74 = 34<\/h3>\n

(vii)\u00a0Given that,\u00a0a\u00a0= 8,\u00a0an\u00a0= 62,\u00a0Sn\u00a0= 210
\nAs, sum of n terms in an AP is,
\nSn\u00a0=\u00a0n\/2 (a\u00a0+\u00a0an)
\n210 =\u00a0n\/2 (8 +62)
\n\u21d2 35n\u00a0= 210
\n\u21d2\u00a0n\u00a0= 210\/35 = 6
\nNow, 62 = 8+5d
\n\u21d2 5d\u00a0= 62-8 = 54
\n\u21d2\u00a0d\u00a0= 54\/5 = 10.8<\/h3>\n

(viii)\u00a0Given that,\u00a0nth\u00a0term,\u00a0an\u00a0= 4,\u00a0common difference,\u00a0d\u00a0= 2,\u00a0sum of n terms,\u00a0Sn\u00a0= \u221214.
\nAs\u00a0we know, from the formula of the nth\u00a0term in an AP,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d,
\nTherefore, putting the given values, we get,
\n4 =\u00a0a+(n\u00a0\u22121)2
\n4 =\u00a0a+2n\u22122
\na+2n\u00a0= 6
\na\u00a0= 6 \u2212 2n\u00a0\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nAs we know, the sum of n terms is;
\nSn\u00a0=\u00a0n\/2 (a+an)
\n-14 =\u00a0n\/2 (a+4)
\n\u221228 =\u00a0n\u00a0(a+4)
\n\u221228 =\u00a0n\u00a0(6 \u22122n\u00a0+4) {From equation\u00a0(i)}
\n\u221228 =\u00a0n\u00a0(\u2212 2n\u00a0+10)
\n\u221228 = \u2212 2n2+10n
\n2n2\u00a0\u221210n\u00a0\u2212 28 = 0
\nn2\u00a0\u22125n\u00a0\u221214 = 0
\nn2\u00a0\u22127n+2n\u00a0\u221214 = 0
\nn\u00a0(n\u22127)+2(n\u00a0\u22127) = 0
\n(n\u00a0\u22127)(n\u00a0+2) = 0
\nEither\u00a0n\u00a0\u2212 7 = 0 or\u00a0n\u00a0+ 2 = 0
\nn\u00a0= 7 or\u00a0n\u00a0= \u22122
\nHowever,\u00a0n\u00a0can neither be negative nor fractional.
\nTherefore,\u00a0n\u00a0= 7
\nFrom equation\u00a0(i), we get
\na\u00a0= 6\u22122n
\na\u00a0= 6\u22122(7)
\n= 6\u221214
\n= \u22128<\/h3>\n

(ix) Given that,\u00a0first term,\u00a0a\u00a0= 3,
\nNumber of terms,\u00a0n\u00a0= 8
\nAnd\u00a0sum of n terms,\u00a0S\u00a0= 192
\nAs we know,
\nSn\u00a0=\u00a0n\/2\u00a0[2a+(n\u00a0-1)d]
\n192 = 8\/2\u00a0[2\u00d73+(8\u00a0-1)d]
\n192 = 4[6 +7d]
\n48 = 6+7d
\n42 = 7d
\nd =\u00a06<\/h3>\n

(x) Given that,\u00a0l\u00a0= 28,S\u00a0= 144 and there are total of 9 terms.
\nSum of n terms formula,
\nSn\u00a0=\u00a0n\/2 (a\u00a0+\u00a0l)
\n144 = 9\/2(a+28)
\n(16)\u00d7(2) =\u00a0a+28
\n32 =\u00a0a+28
\na\u00a0= 4<\/h3>\n

4. How many terms of the AP. 9, 17, 25 \u2026 must be taken to give a sum of 636?<\/h2>\n

Solutions:
\nLet there be\u00a0n\u00a0terms of the AP. 9, 17, 25 \u2026
\nFor this A.P.,
\nFirst term,\u00a0a\u00a0= 9
\nCommon difference, d\u00a0=\u00a0a2\u2212a1\u00a0= 17\u22129 = 8
\nAs, the sum of n terms, is;
\nSn\u00a0=\u00a0n\/2\u00a0[2a+(n\u00a0-1)d]
\n636 =\u00a0n\/2\u00a0[2\u00d7a+(8-1)\u00d78]
\n636 =\u00a0n\/2\u00a0[18+(n-1)\u00d78]
\n636 =\u00a0n\u00a0[9 +4n\u00a0\u22124]
\n636 =\u00a0n\u00a0(4n\u00a0+5)
\n4n2\u00a0+5n\u00a0\u2212636 = 0
\n4n2\u00a0+53n\u00a0\u221248n\u00a0\u2212636 = 0
\nn\u00a0(4n\u00a0+ 53)\u221212 (4n\u00a0+ 53) = 0
\n(4n\u00a0+53)(n\u00a0\u221212) = 0
\nEither 4n+53 = 0 or\u00a0n\u221212 = 0
\nn\u00a0= (-53\/4) or\u00a0n\u00a0= 12
\nn\u00a0cannot be negative or fraction, therefore,\u00a0n\u00a0= 12 only.<\/h3>\n

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.<\/h2>\n

Solution:
\nGiven that,
\nfirst term, a\u00a0= 5
\nlast term, l\u00a0= 45
\nSum of the AP, Sn\u00a0= 400
\nAs we know, the sum of AP formula is;
\nSn\u00a0=\u00a0n\/2 (a+l)
\n400 =\u00a0n\/2(5+45)
\n400 =\u00a0n\/2(50)
\nNumber of terms, n\u00a0=16
\nAs we know, the last term of AP series can be written as;
\nl = a+(n\u00a0\u22121)d
\n45 = 5 +(16 \u22121)d
\n40 = 15d
\nCommon difference, d\u00a0= 40\/15 = 8\/3<\/h3>\n

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?<\/h2>\n

Solution:
\nGiven that,
\nFirst term, a\u00a0= 17
\nLast term, l\u00a0= 350
\nCommon difference, d\u00a0= 9
\nLet there be\u00a0n\u00a0terms in the A.P., thus the formula for last term can be written as;
\nl = a+(n\u00a0\u22121)d
\n350 = 17+(n\u00a0\u22121)9
\n333 = (n\u22121)9
\n(n\u22121) = 37
\nn\u00a0= 38
\nSn\u00a0=\u00a0n\/2 (a+l)
\nS38\u00a0= 38\/2 (17+350)
\n= 19\u00d7367
\n= 6973
\nThus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.<\/h3>\n

7. Find the sum of first 22 terms of an AP in which\u00a0d\u00a0= 7 and 22nd\u00a0term is 149.<\/h2>\n

Solution:
\nGiven,<\/h3>\n

Common difference, d\u00a0= 7
\n22nd\u00a0term, a22\u00a0= 149
\nSum of first 22 term, S22\u00a0= ?
\nBy the formula of nth term,
\nan\u00a0=\u00a0a+(n\u22121)d
\na22\u00a0=\u00a0a+(22\u22121)d
\n149 =\u00a0a+21\u00d77
\n149 =\u00a0a+147
\na\u00a0= 2 = First term
\nSum of n terms,
\nSn\u00a0=\u00a0n\/2(a+an)
\nS22\u00a0= 22\/2 (2+149)
\n= 11\u00d7151
\n= 1661<\/h3>\n

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.<\/h2>\n

Solution:
\nGiven that,
\nSecond term, a2\u00a0= 14
\nThird term, a3\u00a0= 18
\nCommon difference, d\u00a0=\u00a0a3\u2212a2\u00a0= 18\u221214 = 4
\na2\u00a0=\u00a0a+d
\n14 =\u00a0a+4
\na\u00a0= 10 = First term
\nSum of n terms;
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+ (n\u00a0\u2013 1)d]
\nS51\u00a0= 51\/2\u00a0[2\u00d710\u00a0(51-1) 4]
\n= 51\/2 [20+(50)\u00d74]
\n= 51 \u00d7 220\/2
\n= 51 \u00d7 110
\n= 5610<\/h3>\n

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first\u00a0n\u00a0terms.<\/h2>\n

Solution:
\nGiven that,
\nS7\u00a0= 49
\nS17\u00a0= 289
\nWe know, Sum of n terms;
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+ (n\u00a0\u2013 1)d]
\nTherefore,
\nS7=\u00a07\/2\u00a0[2a\u00a0+(n\u00a0-1)d]
\nS7\u00a0= 7\/2\u00a0[2a\u00a0+ (7 -1)d]
\n49 = 7\/2\u00a0[2a\u00a0+6d]
\n7 = (a+3d)
\na\u00a0+ 3d\u00a0= 7 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(i)
\nIn the same way,
\nS17\u00a0= 17\/2\u00a0[2a+(17-1)d]
\n289 = 17\/2 (2a\u00a0+16d)
\n17 = (a+8d)
\na\u00a0+8d\u00a0= 17 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.\u00a0(ii)
\nSubtracting equation\u00a0(i)\u00a0from equation\u00a0(ii),
\n5d\u00a0= 10
\nd\u00a0= 2
\nFrom equation\u00a0(i), we can write it as;
\na+3(2) = 7
\na+\u00a06 = 7
\na =\u00a01
\nHence,
\nSn\u00a0=\u00a0n\/2[2a+(n-1)d]
\n=\u00a0n\/2[2(1)+(n\u00a0\u2013 1)\u00d72]
\n=\u00a0n\/2(2+2n-2)
\n=\u00a0n\/2(2n)
\n=\u00a0n2<\/h3>\n

10. Show that\u00a0a1,\u00a0a2\u00a0\u2026 ,\u00a0an\u00a0, \u2026 form an AP where\u00a0an\u00a0is defined as below
\n(i)\u00a0an\u00a0= 3+4n
\n(ii)\u00a0an\u00a0= 9\u22125n
\nAlso find the sum of the first 15 terms in each case.<\/h2>\n

Solutions:
\n(i)\u00a0an\u00a0= 3+4n
\na1\u00a0= 3+4(1) = 7
\na2\u00a0= 3+4(2) = 3+8 = 11
\na3\u00a0= 3+4(3) = 3+12 = 15
\na4\u00a0= 3+4(4) = 3+16 = 19
\nWe can see here, the common difference between the terms are;
\na2\u00a0\u2212\u00a0a1\u00a0= 11\u22127 = 4
\na3\u00a0\u2212\u00a0a2\u00a0= 15\u221211 = 4
\na4\u00a0\u2212\u00a0a3\u00a0= 19\u221215 = 4
\nHence,\u00a0ak\u00a0+ 1\u00a0\u2212\u00a0ak\u00a0is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.
\nNow, we know, the sum of nth term is;
\nSn\u00a0=\u00a0n\/2[2a+(n\u00a0-1)d]
\nS15\u00a0= 15\/2[2(7)+(15-1)\u00d74]
\n= 15\/2[(14)+56]
\n= 15\/2(70)
\n= 15\u00d735
\n= 525<\/h3>\n

(ii)\u00a0an\u00a0= 9\u22125n
\na1\u00a0= 9\u22125\u00d71 = 9\u22125 = 4
\na2\u00a0= 9\u22125\u00d72 = 9\u221210 = \u22121
\na3\u00a0= 9\u22125\u00d73 = 9\u221215 = \u22126
\na4\u00a0= 9\u22125\u00d74 = 9\u221220 = \u221211
\nWe can see here, the common difference between the terms are;
\na2\u00a0\u2212\u00a0a1\u00a0= \u22121\u22124 = \u22125
\na3\u00a0\u2212\u00a0a2\u00a0= \u22126\u2212(\u22121) = \u22125
\na4\u00a0\u2212\u00a0a3\u00a0= \u221211\u2212(\u22126) = \u22125
\nHence,\u00a0ak\u00a0+ 1\u00a0\u2212\u00a0ak\u00a0is same every time. Therefore, this is an A.P. with common difference as \u22125 and first term as 4.
\nNow, we know, the sum of nth term is;
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n-1)d]
\nS15\u00a0= 15\/2[2(4)\u00a0+(15\u00a0-1)(-5)]
\n= 15\/2[8 +14(-5)]
\n= 15\/2(8-70)
\n= 15\/2(-62)
\n= 15(-31)
\n= -465<\/h3>\n

11. If the sum of the first\u00a0n\u00a0terms of an AP is 4n\u00a0\u2212\u00a0n2, what is the first term (that is\u00a0S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th\u00a0and the\u00a0nth\u00a0terms.<\/h2>\n

Solution:
\nGiven that,
\nSn\u00a0= 4n\u2212n2
\nFirst term,\u00a0a\u00a0=\u00a0S1\u00a0= 4(1) \u2212 (1)2\u00a0= 4\u22121 = 3
\nSum of first two terms =\u00a0S2= 4(2)\u2212(2)2\u00a0= 8\u22124 = 4
\nSecond term,\u00a0a2\u00a0=\u00a0S2\u00a0\u2212\u00a0S1\u00a0= 4\u22123 = 1
\nCommon difference, d\u00a0=\u00a0a2\u2212a\u00a0= 1\u22123 = \u22122
\nNth\u00a0term,\u00a0an\u00a0=\u00a0a+(n\u22121)d
\n= 3+(n\u00a0\u22121)(\u22122)
\n= 3\u22122n\u00a0+2
\n= 5\u22122n
\nTherefore,\u00a0a3\u00a0= 5\u22122(3) = 5-6 = \u22121
\na10\u00a0= 5\u22122(10) = 5\u221220 = \u221215
\nHence, the sum of first two terms is 4. The second term is 1.
\nThe 3rd, the 10th, and\u00a0the\u00a0nth\u00a0terms are \u22121, \u221215, and 5 \u2212 2n\u00a0respectively.<\/h3>\n

12. Find the sum of first 40 positive integers divisible by 6.<\/h2>\n

Solution:
\nThe positive integers that are divisible by 6 are 6, 12, 18, 24 \u2026.
\nWe can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.
\na\u00a0= 6
\nd\u00a0= 6
\nS40\u00a0=\u00a0?
\nBy the formula of sum of n terms, we know,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0\u2013 1)d]
\nTherefore, putting n = 40, we get,
\nS40\u00a0= 40\/2\u00a0[2(6)+(40-1)6]
\n= 20[12+(39)(6)]
\n= 20(12+234)
\n= 20\u00d7246
\n= 4920<\/h3>\n

13. Find the sum of first 15 multiples of 8.<\/h2>\n

Solution:
\nThe multiples of 8 are 8, 16, 24, 32\u2026
\nThe series is in the form of AP, having first term as 8 and common difference as 8.
\nTherefore,\u00a0a\u00a0= 8
\nd\u00a0= 8
\nS15\u00a0= ?
\nBy the formula of sum of nth term, we know,
\nSn\u00a0=\u00a0n\/2\u00a0[2a+(n-1)d]
\nS15\u00a0= 15\/2\u00a0[2(8)\u00a0+ (15-1)8]
\n=\u00a015\/2[16 +(14)(8)]
\n=\u00a015\/2[16 +112]
\n= 15(128)\/2
\n= 15 \u00d7 64
\n= 960<\/h3>\n

14. Find the sum of the odd numbers between 0 and 50.<\/h2>\n

Solution:
\nThe odd numbers between 0 and 50 are 1, 3, 5, 7, 9 \u2026 49.
\nTherefore, we can see that these odd numbers are in the form of A.P.
\nHence,
\nFirst term, a\u00a0= 1
\nCommon difference, d\u00a0= 2
\nLast term,\u00a0l\u00a0= 49
\nBy the formula of last term, we know,
\nl\u00a0=\u00a0a+(n\u22121)\u00a0d
\n49 = 1+(n\u22121)2
\n48 = 2(n\u00a0\u2212 1)
\nn\u00a0\u2212 1 = 24
\nn\u00a0= 25 = Number of terms
\nBy the formula of sum of nth term, we know,
\nSn\u00a0=\u00a0n\/2(a\u00a0+l)
\nS25\u00a0= 25\/2 (1+49)
\n= 25(50)\/2
\n=(25)(25)
\n= 625<\/h3>\n

15.\u00a0A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.<\/h2>\n

Solution:
\nWe can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.
\nTherefore, a\u00a0= 200 and\u00a0d\u00a0= 50
\nPenalty that has to be paid if contractor has delayed the work by 30 days =\u00a0S30
\nBy the formula of sum of nth term, we know,
\nSn\u00a0=\u00a0n\/2[2a+(n\u00a0-1)d]
\nTherefore,
\nS30= 30\/2[2(200)+(30 \u2013 1)50]
\n= 15[400+1450]
\n= 15(1850)
\n= 27750
\nTherefore, the contractor has to pay Rs 27750 as penalty.<\/h3>\n

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.<\/h2>\n

Solution:
\nLet the cost of 1st\u00a0prize be\u00a0Rs.\u00a0P.
\nCost of 2nd\u00a0prize =\u00a0Rs.\u00a0P\u00a0\u2212 20
\nAnd cost of 3rd\u00a0prize =\u00a0Rs.\u00a0P\u00a0\u2212 40
\nWe can see that the cost of these prizes are in the form of A.P., having common difference as \u221220 and first term as\u00a0P.
\nThus, a\u00a0=\u00a0P and d\u00a0= \u221220
\nGiven that,\u00a0S7\u00a0= 700
\nBy the formula of sum of nth term, we know,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+ (n\u00a0\u2013 1)d]
\n7\/2\u00a0[2a\u00a0+ (7 \u2013 1)d]\u00a0= 700<\/h3>\n

a\u00a0+ 3(\u221220) = 100
\na\u00a0\u221260 = 100
\na\u00a0= 160
\nTherefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.<\/h3>\n

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?<\/h2>\n

Solution:
\nIt can be observed that the number of trees planted by the students is in an AP.
\n1, 2, 3, 4, 5\u2026\u2026\u2026\u2026\u2026\u2026..12
\nFirst term,\u00a0a\u00a0= 1
\nCommon difference,\u00a0d\u00a0= 2\u22121 = 1
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n-1)d]
\nS12\u00a0= 12\/2\u00a0[2(1)+(12-1)(1)]
\n= 6(2+11)
\n= 6(13)
\n= 78
\nTherefore, number of trees planted by 1 section of the classes = 78
\nNumber of trees planted by 3 sections of the classes = 3\u00d778 = 234
\nTherefore, 234 trees will be planted by the students.<\/h3>\n

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, \u2026\u2026\u2026 as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take\u00a0\u03c0 = 22\/7)<\/h2>\n

\"\"
\nSolution:
\nWe know,
\nPerimeter of a semi-circle = \u03c0r
\nTherefore,
\nP1\u00a0= \u03c0(0.5) = \u03c0\/2 cm
\nP2\u00a0= \u03c0(1) = \u03c0 cm
\nP3\u00a0= \u03c0(1.5) = 3\u03c0\/2 cm
\nWhere, P1,\u00a0P2,\u00a0P3\u00a0are the lengths of the semi-circles.
\nHence we got a series here, as,
\n\u03c0\/2, \u03c0, 3\u03c0\/2, 2\u03c0, \u2026.
\nP1\u00a0= \u03c0\/2 cm
\nP2\u00a0= \u03c0 cm
\nCommon difference, d\u00a0=\u00a0P2\u00a0\u2013 P1\u00a0=\u00a0\u03c0 \u2013 \u03c0\/2 = \u03c0\/2
\nFirst term =\u00a0P1=\u00a0a\u00a0=\u00a0\u03c0\/2 cm
\nBy the sum of n term formula, we know,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+ (n\u00a0\u2013 1)d]
\nTherefor, Sum of the length of 13 consecutive circles is;
\nS13\u00a0=\u00a013\/2\u00a0[2(\u03c0\/2)\u00a0+ (13\u00a0\u2013 1)\u03c0\/2]
\n= \u00a013\/2\u00a0[\u03c0\u00a0+ 6\u03c0]
\n=13\/2\u00a0(7\u03c0)
\n=\u00a013\/2 \u00d7 7 \u00d7\u00a022\/7
\n= 143 cm<\/h3>\n

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?<\/h2>\n

\"\"
\nSolution:
\nWe can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18\u2026
\nFor the given A.P.,
\nFirst term, a\u00a0= 20 and common difference,\u00a0d\u00a0=\u00a0a2\u2212a1\u00a0= 19\u221220 = \u22121
\nLet a total of 200 logs be placed in\u00a0n\u00a0rows.
\nThus, Sn\u00a0= 200
\nBy the sum of nth term formula,
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0-1)d]
\nS12\u00a0= 12\/2\u00a0[2(20)+(n\u00a0-1)(-1)]
\n400 =\u00a0n\u00a0(40\u2212n+1)
\n400 =\u00a0n\u00a0(41-n)
\n400 = 41n\u2212n2
\nn2\u221241n\u00a0+ 400 = 0
\nn2\u221216n\u221225n+400 = 0
\nn(n\u00a0\u221216)\u221225(n\u00a0\u221216) = 0
\n(n\u00a0\u221216)(n\u00a0\u221225) = 0
\nEither (n\u00a0\u221216) = 0 or\u00a0n\u221225 = 0
\nn\u00a0= 16 or\u00a0n\u00a0= 25
\nBy the nth term formula,
\nan\u00a0=\u00a0a+(n\u22121)d
\na16\u00a0= 20+(16\u22121)(\u22121)
\na16\u00a0= 20\u221215
\na16\u00a0= 5
\nSimilarly, the 25th\u00a0term could be written as;
\na25\u00a0= 20+(25\u22121)(\u22121)
\na25\u00a0= 20\u221224
\n= \u22124
\nIt can be seen, the number of logs in 16th\u00a0row is 5 as the numbers cannot be negative.
\nTherefore, 200 logs can be placed in 16 rows and the number of logs in the 16th\u00a0row is 5.<\/h3>\n

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.<\/h2>\n

\"\"
\nA competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
\n[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2\u00d75+2\u00d7(5+3)]<\/h2>\n

Solution:
\nThe distances of potatoes from the bucket are 5, 8, 11, 14\u2026, which is in the form of AP.
\nGiven, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.
\nTherefore, distances to be run w.r.t distances of potatoes, could be written as;
\n10, 16, 22, 28, 34,\u2026\u2026\u2026.
\nHence, the first term, a\u00a0= 10 and\u00a0d\u00a0= 16\u221210 = 6
\nS10\u00a0=?
\nBy the formula of sum of n terms, we know,
\nS10\u00a0= 10\/2\u00a0[2(10)+(10\u00a0-1)(6)]
\n= 5[20+54]
\n= 5(74)
\n= 370
\nTherefore, the competitor will run a total distance of 370 m.<\/h3>\n

Exercise 5.4<\/h2>\n

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an\u00a0< 0]<\/h2>\n

Solution:
\nGiven the AP series is 121, 117, 113, . . .,
\nThus, first term, a = 121
\nCommon difference, d = 117-121= -4
\nBy the nth term formula,
\nan\u00a0=\u00a0a+(n\u00a0\u22121)d
\nTherefore,
\nan\u00a0=\u00a0121+(n\u22121)(-4)
\n= 121-4n+4
\n=125-4n
\nTo find the first negative term of the series,\u00a0an\u00a0< 0
\nTherefore,
\n125-4n < 0
\n125 < 4n n>125\/4
\nn>31.25
\nTherefore,\u00a0the first negative term of the series is 32nd\u00a0term.<\/h3>\n

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.<\/h2>\n

Solution:
\nFrom the given statements, we can write,
\na3 +\u00a0a7\u00a0= 6 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(i)
\nAnd
\na3\u00a0\u00d7a7\u00a0= 8 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(ii)
\nBy the nth term formula,
\nan\u00a0=\u00a0a+(n\u22121)d
\nThird term,\u00a0a3\u00a0= a+(3 -1)d
\na3\u00a0= a + 2d\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(iii)
\nAnd Seventh term, a7= a+(7-1)d
\na7\u00a0= a + 6d \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(iv)
\nFrom equation (iii) and (iv), putting in equation(i), we get,
\na+2d +a+6d = 6
\n2a+8d = 6
\na+4d=3
\nor
\na = 3\u20134d \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(v)
\nAgain putting the eq.(iii) and (iv), in eq. (ii), we get,
\n(a+2d)\u00d7(a+6d) = 8
\nPutting the value of a from equation (v), we get,
\n(3\u20134d +2d)\u00d7(3\u20134d+6d) = 8
\n(3 \u20132d)\u00d7(3+2d) = 8
\n32\u00a0\u2013 2d2\u00a0= 8
\n9 \u2013 4d2\u00a0= 8
\n4d2\u00a0= 1
\nd = 1\/2 or -1\/2
\nNow, by putting both the values of d, we get,
\na = 3 \u2013 4d = 3 \u2013 4(1\/2) = 3 \u2013 2 = 1, when d = 1\/2
\na = 3 \u2013 4d = 3 \u2013 4(-1\/2) = 3+2 = 5, when d = -1\/2
\nWe know, the sum of nth term of AP is;
\nSn\u00a0=\u00a0n\/2\u00a0[2a\u00a0+(n\u00a0\u2013 1)d]
\nSo, when a = 1 and d=1\/2
\nThen, the sum of first 16 terms are;
\nS16\u00a0=\u00a016\/2\u00a0[2\u00a0+(16-1)1\/2] = 8(2+15\/2) = 76
\nAnd when a = 5 and d= -1\/2
\nThen, the sum of first 16 terms are;
\nS16\u00a0=\u00a016\/2\u00a0[2(5)+(16-1)(-1\/2)] = 8(5\/2)=20<\/h3>\n

3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are<\/h2>\n

\"\"
\napart, what is the length of the wood required for the rungs? [Hint: Number of rungs = -250\/25 ].<\/h2>\n

\"\"
\nSolution:
\nGiven,
\nDistance between the rungs of the ladder is 25cm.
\nDistance between the top rung and bottom rung of the ladder is =<\/h3>\n

\"\"
\n= 5\/2 \u00d7100cm
\n= 250cm
\nTherefore, total number of rungs = 250\/25 + 1 = 11
\nAs we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.
\nAnd the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.
\nSo,
\nFirst term, a = 45
\nLast term,\u00a0l =\u00a025
\nNumber of terms, n = 11
\nNow, as we know, sum of nth terms is equal to,
\nSn= n\/2(a+\u00a0l)
\nSn= 11\/2(45+25) = 11\/2(70) = 385 cm
\nHence, the length of the wood required for the rungs is 385cm.<\/h3>\n

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx \u2013 1 = S49 \u2013 Sx ]<\/h2>\n

Solution:
\nGiven,
\nRow houses are numbers from 1,2,3,4,5\u2026\u2026.49.
\nThus we can see the houses numbered in a row are in the form of AP.
\nSo,
\nFirst term, a = 1
\nCommon difference, d=1
\nLet us say the number of xth\u00a0houses can be represented as;
\nSum of nth term of AP = n\/2[2a+(n-1)d]
\nSum of number of houses beyond x house = Sx-1
\n= (x-1)\/2[2(1)+(x-1-1)1]
\n= (x-1)\/2 [2+x-2]
\n= x(x-1)\/2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(i)
\nBy the given condition, we can write,
\nS49\u00a0\u2013 Sx\u00a0= {49\/2[2(1)+(49-1)1]}\u2013{x\/2[2(1)+(x-1)1]}
\n= 25(49) \u2013 x(x + 1)\/2 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(ii)
\nAs per the given condition, eq.(i) and eq(ii) are equal to each other;
\nTherefore,
\nx(x-1)\/2 = 25(49) \u2013 x(x+1)\/2
\nx = \u00b135
\nAs we know, the number of houses cannot be a negative number. Hence, the value of x is 35.<\/h3>\n

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1 4 m and a tread of 1 2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = \u00bc \u00d71\/2 \u00d750 m3.]<\/h2>\n

\"\"
\nSolution:
\nAs we can see from the given figure, the first step is \u00bd m wide, 2nd\u00a0step is 1m wide and 3rd\u00a0step is 3\/2m wide. Thus we can understand that the width of step by \u00bd m each time when height is \u00bc m. And also, given length of the steps is 50m all the time. So, the width of steps forms a series AP in such a way that;
\n\u00bd , 1, 3\/2, 2, \u2026\u2026..
\nVolume of steps = Volume of Cuboid
\n= Length \u00d7 Breadth Height
\nNow,
\nVolume of concrete required to build the first step = \u00bc \u00d71\/2 \u00d750 = 25\/4
\nVolume of concrete required to build the second step =\u00bc \u00d71\u00d750 = 25\/2
\nVolume of concrete required to build the second step = \u00bc \u00d73\/2 \u00d750 = 75\/4
\nNow, we can see the volumes of concrete required to build the steps, are in AP series;
\n25\/4 , 25\/2 , 75\/4 \u2026..
\nThus, applying the AP series concept,
\nFirst term, a = 25\/4
\nCommon difference, d = 25\/2 \u2013 25\/4 = 25\/4
\nAs we know, the sum of n terms is;
\nSn\u00a0= n\/2[2a+(n-1)d] = 15\/2(2\u00d7(25\/4 )+(15\/2 -1)25\/4)
\nUpon solving, we get,
\nSn =\u00a015\/2 (100)
\nSn=750
\nHence, the total volume of concrete required to build the terrace is 750 m3.<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 5 – Arithmetic Progressions Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119094","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119094","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119094"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119094\/revisions"}],"predecessor-version":[{"id":119138,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119094\/revisions\/119138"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119094"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119094"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119094"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}