Three numbers are in AP, and their sum is 21.

Solution:

To find: Three numbers

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

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

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

According to first condition

$a+d+a+a-d=21$

$\Rightarrow 3 a=21$

$\Rightarrow a=7$

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 –

$7-d, 7-1,7+d+1$

$\Rightarrow 7-d, 6,8+d$

The above numbers are in GP

Therefore, $6^{2}=(7-d)(8+d)$

$\Rightarrow 36=56+7 d-8 d-d^{2}$

$\Rightarrow d^{2}+d-20=0$

$\Rightarrow d^{2}+5 d-4 d-20=0$

$\Rightarrow d(d+5)-4(d+5)=0$

$\Rightarrow(d-4)(d+5)=0$

$\Rightarrow d=4$, 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.