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, b2 = 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, 62 = (7 – d) (8 + d)

⇒ 36 = 56 + 7d – 8d – d2

⇒ d2 + d – 20 = 0

⇒ d2 + 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.