The sum of three consecutive terms of an AP is 21, and the sum of the

1.	The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms
Answer» To Find: The three numbers which are in AP. Given: Sum and sum of the squares of three numbers are 21 and 165 respectively. Let required number be (a - d), (a), (a + d). Then, (a – d) + a + (a + d) = 21 ⇒ 3a = 21 ⇒ a = 7 Thus, the numbers are (7 - d), 7 and (7 + d). But their sum of the squares of three numbers is 165. ∴ (7 - d)² + 7² + (7 + d)²= 165 ⇒ 49 + d² –14d + 49 + d²+ 14d = 116 ⇒ 2d² = 18 ⇒ d² = 9 ⇒ d = ± 3 When d = 3 numbers are 4, 7, 10 When d = (– 3) numbers are 10, 7, 4 So, Numbers are 4, 7, 10 or 10, 7, 4.

The sum of three consecutive terms of an AP is 21, and the sum of the squares of these terms is 165. Find these terms

Answer»

To Find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let required number be (a - d), (a), (a + d).

Then,

(a – d) + a + (a + d) = 21

⇒ 3a = 21

⇒ a = 7

Thus, the numbers are (7 - d), 7 and (7 + d).

But their sum of the squares of three numbers is 165.

∴ (7 - d)² + 7² + (7 + d)²= 165

⇒ 49 + d² –14d + 49 + d²+ 14d = 116

⇒ 2d² = 18

⇒ d² = 9

⇒ d = ± 3

When d = 3 numbers are 4, 7, 10

When d = (– 3) numbers are 10, 7, 4

So, Numbers are 4, 7, 10 or 10, 7, 4.