Question

Find four consecutive term in ap whose sum is 20 and the sum of whose squares is 120

Answer 1

Let the four numbers in A.P be a-3d, a-d,a+d,a+3d. ---- (1)

Given that Sum of the terms = 20.

= (a-3d) + (a-d) + (a+d) + (a+3d) = 20

4a = 20

a = 5. ---- (2)

Given that sum of squares of the term = 120.

= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120

= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120

= 4a^2 + 20d^2 = 120

Substitute a = 5 from (2) .

4(5)^2 + 20d^2 = 120

100 + 20d^2 = 120

20d^2 = 20

d = +1 (or) - 1.

Since AP cannot be negative.

Substitute a = 5 and d = 1 in (1), we get

a - 3d, a-d, a+d, a+3d = 2,4,6,8.

Hope this helps!