Find b to make the three numbers in AP For a positive number '$b$', the value of '$b$' for which the numbers $3^x + 3^{-x} , b ,9^x+ 9^{-x}$ are in A.P. can be : (A) 1 (B) 2 (C) 3 (D) 5​
Hello! This question has multiple correct answers and belongs to the topic Arithmetic Progression(Sequence and series). I was able to get the answer b=2. As we know that any number of the form $y + y^{-1}$ can take all values in the set = $(- ∞,-2]∪[2,∞)$. By taking $3^x + 3^{-x} = 2$, I got a constant AP = $2,2,2$. The answers are option (B),(C) and (D). Please help me reach the full solution.

\begin{align}
\frac{3^x+3^{-x}+9^x + 9^{-x}}{2} &= \frac{3^x+3^{-x}+(3^x+3^{-x})^2-2}{2}\\
&= -1 + \frac{(3^x+3^{-x})+(3^x+3^{-x})^2}{2}
\end{align}
We know that the range of $3^{x}+3^{-x}$ is $[2, \infty)$
Hence the range of $(3^x+3^{-x})+(3^x+3^{-x})^2$ is  $[6, \infty)$.
Hence $b$ can attain any number at least $\frac{6}{2}-1=2$.
