+1632 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
+118687 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
