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

CONCEPT:

Let us suppose a be the first term and d be the common difference of an AP. Then the nth term of an AP is given by:an = a + (n - 1) × d.

Note: If l is the last term of a sequence, then l = an = a + (n - 1) × d.

CALCULATION:

Given: The sum of three consecutive terms of an AP is 21 and the sum of the squares of these numbers is 165.

Let (a - d), a, (a + d) be the required three consecutive terms of an AP.

∵ Sum of these three numbers is 21.

i.e (a - d) + a + (a + d) = 21

⇒ 3a = 21

⇒ a = 7.

∵ Sum of their squares is 165.

i.e (a - d)² + a² + (a + d)² = 165

⇒ 3a² + 2d² = 165---------(1)

By substituting a = 7 in equation (1), we get

⇒ 3 ⋅ (7)² + 2 ⋅ d² = 165

⇒ 147 + 2 ⋅ d² = 165

⇒ 2 ⋅ d2 = 18 

⇒ d = ± 3

Case 1: When a = 7 and d = 3 the three terms are: 4, 7, 10

Case 2: When a = 7 and d = - 3 the three terms are: 10, 7, 4

Hence, in either case the three numbers are 4, 7, 10