Question :The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. The correct answer :Let there be n terms in the arithmetic progression (AP), and let the common difference be d. Then, we have: The nth term = a + (n-1)d where a is the first term. According to the problem, we have: a = 5 (first term) a + (n-1)d = 45 (last term) n/2 [2a + (n-1)d] = 400 (sum of n terms) Substituting a = 5 and a + (n-1)d = 45, we get: 5 + (n-1)d = 45 n = 41/d + 1 Substituting these values into the third equation, we get: n/2 [10 + (n-1)d] = 400 Substituting n = 41/d + 1, we get: (41/d + 1)/2 [10 + (41/d) d - d] = 400 Simplifying, we get: (41/d + 1)/2 [10 + 40] = 400 Multiplying both sides by 2, we get: (41/d + 1) x 25 = 800 Expanding, we get: 41/d + 1 = 32 Subtracting 1 from both sides, we get: 41/d = 31 Multiplying both sides by d, we get: d = 41/31 Substituting d = 41/31 into the equation n = 41/d + 1, we get: n = 41/(41/31) + 1 = 32 Therefore, the common difference is d = 41/31 and the number of terms is n = 32.
