Question

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

Answer 1

Answer = 2,4,6,8

Step-by-step explanation:

Suppose the four numbers in A.P

a - 3d , a - d , a + d , a + 3d. ... (Eqn 1)

According to the Question

Sum of the terms = 20.

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

4a = 20

a = 5.    ......  (Eqn 2)

According to the Question

Sum of squares of the term is 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

Put the value a = 5 from (2) .

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

100 + 20d^2 = 120

20d^2 = 20

d = ±1

AP cannot be negative.

Put the value a = 5 and d = 1 in (1), we get

a - 3d , a - d, a + d, a + 3d

= 2,4,6,8