Sequences and Series: Geometric Progressions

Let $r$ be the common ratio and $a_n$ a general term, then $a_n=7 \cdot r^{n-1}.$ Since we are given $\displaystyle{a_6 = \frac{7}{32}},$ it follows that

$\begin{aligned} 7 r^5 &= \frac{7}{32} \\ r^5 & = \frac{1}{32} \\ r & = \frac{1}{2}. \end{aligned}$

Thus, the third term $k_2$ is

$\begin{aligned} a_3 & =7 \cdot \left(\frac{1}{2}\right)^2 \\ & =\frac{7}{4}. \end{aligned}$