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

Let the 3 consecutive terms are
$a - d, a, a + d\,\,\,\, d > 0$
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
$2 d^2=32 \Rightarrow d^2=16 \Rightarrow d=4 (d=-4 \text { rejected }) $
$\text { Hence from (4) } $
$4(2 a -4)=264 \Rightarrow 2 a -4=66 \Rightarrow 2 a =70 \Rightarrow a =35$
$\therefore K =35^2-300=1225-300=925 $