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.

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.