The third term of the arithmetic progression is 5, and the sum of the first ten terms is 75.

The third term of the arithmetic progression is 5, and the sum of the first ten terms is 75. Find the sum of the squares of the second and fourth terms of this arithmetic progression?

Let’s denote the 1st term of the progression through n1, and the difference of the progression through d.Then the 3rd term will be equal to:

n3 = n1 + 2 * q = 5.p1 = 5 – 2 * q.

The sum of ten members of the progression is:

(n1 + n10) * 10/2 = 5 * (n1 + n1 + 9 * d) = 75;

2 * n1 + 9 * d = 15.2 * (5 – 2 * d) + 9 * d = 10 – 4 * d + 9 * d = 15.

5 * d = 5; q = 5: 5 = 1.

n1 = 5 – 2 * q = 5 – 2 * 1 = 3.

Let’s find the second and fourth terms of the progression:

n2 = n1 + q = 3 + 1 = 4; n4 = n1 + 3 * q = 3 + 3 * 1 = 6.

(n2) ^ 2 + (n4) ^ 2 = 4 ^ 2 + 6 ^ 2 = 16 + 36 = 52.

Answer: 52.