Three numbers are in AP, and their sum is 15. If 1, 4, 19

Solution:

To find: The numbers

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

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 = 15

$\Rightarrow 3 a=15$

$\Rightarrow a=5$

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, $9^{2}=(6-d)(24+d)$

$\Rightarrow 81=144-24 d+6 d-d^{2}$

$\Rightarrow 81=144-18 d-d^{2}$

$\Rightarrow d^{2}+18 d-63=0$

$\Rightarrow d^{2}+21 d-3 d-63=0$

$\Rightarrow d(d+21)-3(d+21)=0$

$\Rightarrow(d-3)(d+21)=0$

$\Rightarrow d=3$, 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.