Find the sum of the first forty terms of the arithmetic progression (An), if A7 = 6, A17 = 26

Let us express the seventh and seventeenth terms of the progression in terms of the first term and the difference:

a7 = a1 + 6d;

a17 = a1 + 16d.

a1 + 6d = 6;

a1 + 16d = 26.

Subtract the first from the second equation and find d:

10d = 20;

d = 20: 10;

d = 2.

a1 + 2 * 6 = 6;

a1 = 6 – 12 = – 6.

Using the formula Sn = (2a1 + d (n – 1)) * n / 2, we find S40:

S40 = (2 * (- 6) + 2 * (40 – 1)) * 40/2 = (- 12 + 78) * 20 = 66 * 20 = 1320.

Answer: S40 = 1320.