Question

Three numbers form an arithmetic progression. If we add $8$ to the first number, we get a geometric progression with the sum of the terms equal to $26$. Find the numbers.

Answer 1

let ap be a-b,a,a+b

and a-b+8,a,a+b are in gp

$⟹a-b+8+a+a+b=26$

$3a+8=26$

$3a=18$

$a=6$

$⟹a^2=(a-b+8)(a+b)$

$36=(14-b)(6+b)$

$36+b^2+8b-84=0$

$b^2+8b-48=0$

$b=4;-12$

numbers are $10,6,2$ $\because$ another set dosn't make gp on addindg 8