Arithmetic and Geometric Sequences Together

Level pending

Suppose x , y , z x,y,z is a geometric sequence with a common ratio of r r where r 1 r \neq 1 . If 3 x , 26 y , 17 z 3x, 26y, 17z is an arithmetic sequence, then the sum of all possible values of r r can be expressed as m n \dfrac{m}{n} . Find m + 2 n m+2n .


The answer is 86.

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2 solutions

Hahn Lheem
Mar 30, 2014

Since x x , y y , and z z are in a geometric sequence, we can express y y as x r xr and z z as x r 2 xr^2 . Therefore, since 3 x 3x , 26 y 26y , and 17 z 17z are in an arithmetic progression, we can construct the equation 17 z 26 y = 26 y 3 x 17z-26y=26y-3x . We can plug in the values of y y and z z in the equation to get 17 x r 2 26 x r = 26 x r 3 x . 17xr^2-26xr=26xr-3x. We can eliminate the x x terms to get 17 r 2 26 r = 26 r 3. 17r^2-26r=26r-3. From this, we can rearrange terms and get the quadratic 17 r 2 52 r + 3 = 0. 17r^2-52r+3=0. By Vieta's, the sum of the roots of a quadratic is equivalent to b a -\dfrac{b}{a} , so the sum of all possible values of r r is 52 17 \dfrac{52}{17} . Our answer is 52 + 2 ( 17 ) = 86 52+2(17)=\boxed{86} .

wat da fuc

Am Kemplin - 1 month, 3 weeks ago

5+5+5+55++5+55++5+5+5+55+5+5+5+3+3+3+3+4+4+4+4+3+4+43+3+4+4+4+4+3+3+6+6+7+7+2+233-2+3+-3+++3-4=100

Am Kemplin - 1 month, 3 weeks ago
Aayush Patni
Nov 27, 2014

I did the same.

fuck this question

Am Kemplin - 1 month, 3 weeks ago

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