Question

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

Answer 1

Sol:
Let a - 3d ,a - d , a + d , a + 3d are 4 consecutive terms in A.P whose common difference is 2d.

Given that sum of 4 consecutive terms in A.P is 20.

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

∴ a = 5

Sum of whose squares is 120.

⇒ (a - 3d)2 + (a - d)2 + (a + d)2 + (a + 3d)2 = 120.

⇒ (5 - 3d)2 + (5 - d)2 + (5 + d)2 + (5 + 3d)2 = 120.

⇒ 2(52 + (3d)2 )+ 2(52 + (d)2 ) = 120.

⇒ 50 + 18d2 + 50 + 2d2  = 120.

⇒ 20d2 = 120 -100.

⇒  d2 = 1

∴ d = ± 1.

Now a = 5 and d = 1

Four consecutive terms in A.P are a - 3d ,a - d , a + d , a + 3d

∴ 2, 4 , 6, 8 are four consecutive terms in A.P.