Find the sum of six terms of an arithmetic progression if a2 = 7 a4 = -1.

First, let’s find the first term a1 and the difference d of this arithmetic sequence.

In the initial data for this task, it is reported that the second term of this progression is 7, and the fourth is -1.

Applying the formula of the member of the arithmetic progression, which is in the n-th position for n = 2 and n = 4, we get:

a1 + d = 7;

a1 + 3d = -1.

We solve the resulting system of equations.

Subtracting the first equation from the second, we get:

a1 + 3d – a1 – d = -1 – 7;

2d = -8;

d = -8 / 2 = -4.

Find a1:

a1 = 7 – d = 7 – (-4) = 7 + 4 = 11.

Applying the formula for the sum of the first n terms of the arithmetic progression with n = 6, we get:

S6 = (2 * a1 + d * (6 – 1)) * 6/2 = (2 * a1 + d * 5) * 3 = (2 * 11 + (-4) * 5) * 3 = (22 – 20) * 3 = 2 * 3 = 6.

Answer: the required amount is 6.