The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165, then what are the terms?

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5, 8, 11 or 11, 8, and 5

4, 7, 10 or 10, 7, and 4

6, 9, 15 or 15, 9, and 6

9, 6, 17 or 17, 6, and 9

answer is B.

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Detailed Solution

Given that the sum of three consecutive terms of an A.P. is 21 and the sum of the squares of these terms is 165.
Let the required terms are

a − d, a

and

a + d

.
As the sum of these terms is 21.

∴ (a − d) + a + (a + d) = 21 ⇒ 3 a = 21 ⇒ a = 213 ⇒ a = 7

As the sum of the square of these terms is 165.

∴ (a − d) 2 + a 2 + (a + d) 2 = 165

The formula used for the subtraction of two variables:

a − b 2 = a 2 + b 2 − 2 a b

.
The formula used for the sum of two variables:

a + b 2 = a 2 + b 2 + 2 a b

Applying the formula of subtraction of two variables and sum of two variables:

⇒ a 2 + d 2 − 2 a d + a 2 + a 2 + d 2 + 2 a d = 165

⇒ 3 a 2 + 2 d 2 = 165

…… (1)
Putting the value of a in equation (1):

⇒ 3 × 72 + 2 d 2 = 165 ⇒ 2 d 2 = 165 − 147 ⇒ d = 182 ⇒ d = ± 3

Take d=3, a=7 and determining three numbers of A.P.:

∴ (7 − 3), 7, and (7 + 3) ⇒ 4, 7, and 10

Take d = -3, a=7 and determining three numbers of A.P.:

∴ 7 − − 3, 7, and 7 + − 3 ⇒ 7 + 3, 7, and 7 − 3 ⇒ 10, 7, and 4

Therefore, the required terms are 4, 7 and 10 or 10, 7, and 4.
Hence, option (2) is correct.

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