Answer : To find: The numbers

Given: Three numbers are in A.P. Their sum is 15 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 = 15

⇒ 3a = 15

⇒ a = 5

Hence numbers are 5 – d, 5, 5 + d

When 1, 4, 19 be added to them respectively then the numbers become – 5 – d + 1, 5 + 4, 5 + d + 19

⇒ 6 – d, 9, 24 + d

The above numbers are in GP Therefore, 92 = (6 – d) (24 + d)

⇒ 81 = 144 – 24d +6d – d2

⇒ 81 = 144 – 18d – d2

⇒ d2 + 18d – 63 = 0

⇒ d2 + 21d – 3d – 63 = 0

⇒ d (d + 21) -3 (d + 21) = 0

⇒ (d – 3) (d + 21) = 0

⇒ d = 3, Or d = -21

Taking d = 3, the numbers are 5 – d, 5, 5 + d = 5 – 3, 5, 5 + 3

= 2, 5, 8

Taking d = -21, the numbers are

5 – d, 5, 5 + d = 5 – (-21), 5, 5 + (-21)

= 26, 5, -16

Ans) We have two sets of triplet as 2, 5, 8 and 26, 5, -16.