The Geometric Squares of an Arithmetic Sequence

Algebra Level 4

If three real numbers are in an arithmetic progression while their squares are in a geometric progression, find the sum of all the possible ratios of the geometric progression of the squares.


The answer is 7.

David Vreken by David Vreken , 42, USA

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1 solution

Let the numbers be a − b , a , a + b a-b, a, a+b . Then a 4 = ( a − b ) 2 ( a + b ) 2 = ( a 2 − b 2 ) 2 a^4=(a-b) ^2(a+b)^2=(a^2-b^2)^2 . Or b 4 = 2 a 2 b 2 b^4=2a^2b^2 , and hence b = 0 b=0 , b = a √ 2 b=a√2 , b = − a √ 2 b=-a√2 . Therefore the common ratios are 1 , 1 3 − 2 √ 2 , 1 3 + 2 √ 2 1, \dfrac{1}{3-2√2}, \dfrac{1}{3+2√2} , and their sum is 1 + 1 3 − 2 √ 2 + 1 3 + 2 √ 2 = 7 1+\dfrac{1}{3-2√2}+\dfrac{1}{3+2√2}=\boxed 7

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This is a good question. I got it at once. :)

Nikola Alfredi - 1 year, 4 months ago

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