1989 AIME Problem 7

Question

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 series. Find

k.

Answer

Write the three arithmetic-sequence terms as

x,

x+r,

and

x+2r

Subtract adjacent square equations, then subtract those two resulting equations

Let the three terms be

x,

x+r,

and

x+2r.

Subtracting the square equations gives

r(2x+r)=300-36=264

and

r(2x+3r)=596-300=296.

Their difference is

2r^2=32,

so

r=\pm4.

Negating all three terms does not change their squares, so take

r=4.

Then

4(2x+4)=264,

giving

x=31.

Therefore

k=x^2-36=961-36=925.

Problem 7 in other years