Answer :

To find: Three numbers

Given: Three numbers are in A.P. Their sum is 21

Formula used: When a,b,c are in GP, b^{2} = ac

Let the numbers be a - d, a, a + d

According to first condition

a + d + a +a – d = 21

⇒ 3a = 21

⇒ a = 7

Hence numbers are 7 - d, 7, 7 + d

When second number is reduced by 1 and third is increased by 1 then the numbers become –

7 – d , 7 – 1 , 7 + d + 1

⇒ 7 – d , 6 , 8 + d

The above numbers are in GP

Therefore, 6^{2} = (7 – d) (8 + d)

⇒ 36 = 56 + 7d – 8d – d^{2}

⇒ d^{2} + d – 20 = 0

⇒ d^{2} + 5d – 4d – 20 = 0

⇒ d (d + 5) – 4 (d + 5) = 0

⇒ (d – 4) (d + 5) = 0

⇒ d = 4, Or d = -5

Taking d = 4, the numbers are

7 - d, 7, 7 + d = 7 - 4, 7, 7 + 4

= 3, 7, 11

Taking d = -5, the numbers are

7 - d, 7, 7 + d = 7 – (-5), 7, 7 + (-5)

= 12, 7, 2

Ans) We have two sets of triplet as 3, 7, 11 and 12, 7, 2.

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