Three numbers are in AP, and their sum is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three numbers in GP. Find the numbers.
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.