Question 1094813: The squares of three positive integers are in arithmetic progression, and the third integer is 12 greater than the first. Find the second integer.
a) 10 b) 14 c) 11 d) 16 e) 7
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52908)   (Show Source):
You can put this solution on YOUR website! .
a) 10.
The numbers are 2, 10 and 14.
14 - 2 = 12.
Their squares are 4, 100, and 196.
The squares form an arithmetic progression.
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
I don't see a workable algebraic solution....
The condition that the squares of three integers a, b, and c are in arithmetic progression, with c being 12 greater than a will be satisfied if we
(1) choose an integer value for a;
(2) set c = a+12;
(3) evaluate a^2 and c^2; and
(4) find the average of a^2 and c^2
If that average is a perfect square, then that average makes an arithmetic sequence with a^2 and c^2; then the second integer we are looking for is the square root of that average.
So try a=1; if that doesn't give you a solution, try a=2; and so on.
a = 1; c = 13; a^2 = 1; c^2 = 169; the average of a^2 and c^2 is 85... not a perfect square.
a = 2; c = 14; a^2 = 4; c^2 = 196; the average of a^2 and c^2 is 100...AHA!
The first and last integers are 2 and 14; the second integer that we are looking for is the square root of 100, which is 10.
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