An algebra problem by Rahil Sehgal

Let the geometric progression be $\left \{\dfrac br, b, br \right \}$ . Then we have:

$\begin{cases} \dfrac br + b + br = b\left(\dfrac 1r+1+r\right) = 13 & ...(1) \\ \dfrac {b^2}{r^2} + b^2 + b^2r^2 = b^2\left(\dfrac 1{r^2}+1+r^2\right) = 91 & ...(2) \end{cases}$

$\begin{aligned} (1): \quad b\left(\frac 1r+1+r\right) & = 13 \\ b\left(\frac 1r+r\right) +b & = 13 \\ b\left(\frac 1r+r\right) & = 13 - b & \small \color{#3D99F6} \text{Squaring both sides} \\ b^2\left(\frac 1{r^2}+2+r^2\right) & = 169 - 26b + b^2 \\ {\color{#3D99F6}b^2\left(\frac 1{r^2}+1+r^2\right)} + b^2 & = 169 - 26b + b^2 & \small \color{#3D99F6} \text{Note that }(2): b^2\left(\frac 1{r^2}+1+r^2\right) = 91 \\ {\color{#3D99F6}91} & = 169 - 26b \\ 26b & = 78 \\ \implies b & = 3 \end{aligned}$

Now, we have:

$\begin{aligned} (1): \quad b\left(\frac 1r+1+r\right) & = 13 \\ 3\left(\frac 1r+1+r\right) & = 13 \\ \frac 3r+3+3r & = 13 \\ \frac 3r-10+3r & = 0 & \small \color{#3D99F6} \text{Multiplying both sides by }r \\ 3r^2 - 10r + 3 & = 0 \\ (3r-1)(r-3) & = 0 \\ \implies r & = 3, \frac 13 \end{aligned}$

Therefore, the geometric progression is $\{1, 3, 9\}$ . The largest number of the geometric progression is $\boxed{9}$ .