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Question: Seven positive numbers form an arithmetic sequence, and their reciprocals form a geometric sequence. If the third term of the arithmetic sequence is 12, what is the greatest possible value of the seventh term?

 

I tried using trial and error to find the possible original sequence, but I am pretty sure you are supposed to do this question with crazy formulas, inequalities, and other shenanigans. Thanks for helping.. 

 

:D

 Feb 21, 2022
 #1
avatar+117224 
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AP:     a, a+d,  a+2d,  a+3d,  .....

 

GP      \(\frac{1}{a},\; \frac{1}{a+d},\;\frac{1}{a+2d}, \dots\)

 

\(r=\frac{T_n}{T_{n-1}}=\frac{1}{a+d}\div\frac{1}{a}=\frac{1}{a+2d}\div\frac{1}{a+d}\;\;etc\\ \frac{a}{a+d}=\frac{a+d}{a+2d}\\ a(a+2d)=(a+d)(a+d)\\ a(a+d)+ad=a(a+d)+d(a+d)\\ ad=ad+d^2\\ d^2=0 \\ d=0\\ \)

 

Hence every term in the AP is 12 and every term in the GP is 1/12

 Feb 21, 2022
 #2
avatar+516 
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Wow! Thanks melody..

 

no wonder why i couldnt guess and check, they were all the same!

 Feb 21, 2022
 #3
avatar+117224 
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You are welcome :)

Melody  Feb 22, 2022

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