Problem:
If the integer k is added to each of the numbers 36,300 and 596, one obtains the squares of three consecutive terms of an arithmetic sequence. Find k.
Subtracting (1) from (3), we find that 4nd=560, from which
nd=140(4)
Multiplying equation (2) by 2 and subtracting the result from the sum of (1) and (3), we find that 2d2=32, giving d=±4. Combining this with (4), we find n=±35. Using (2) again, we have k=n2−300=352−300=925​.