Question

Let $K$ is a positive integer such that $36+K,300+K,596+K$ are the squares of three consecutive terms of an arithmetic progression. Find $K$

Answer 1

Answer: 925

Let the 3 consecutive terms are
$a-d,a,a+dd>0$
hence $a^2-2ad+d^2=36+K$ ....(1)
$a^2=300+K$ ....(2)
$a^2+2ad+d^2=596+K$ ....(3)
now (2) - (1) gives
$d(2a-d)=264$ ....(4)
(3) - (2) gives
$d(2a+d)=296$....(5)
(5)-(4) gives
$2d^2=32\Rightarrow d^2=16\Rightarrow d=4(d=-4\text{ rejected })$
$\text{ Hence from (4) }$
$4(2a-4)=264\Rightarrow 2a-4=66\Rightarrow 2a=70\Rightarrow a=35$
$\therefore K=35^2-300=1225-300=925$