Answer
4
Work Step by Step
The nth partial sum of a geometric sequence
$S_{n}=\displaystyle \sum_{k=1}^{n}ar^{k-1}$ (where $r\neq 1$)
is given by$ \displaystyle \quad S_{n}=a\cdot\frac{1-r^{n}}{1-r}$
---------------------
Given $S_{3}=52, r=3,$ we find a:
$52=a\displaystyle \cdot\frac{1-3^{3}}{1-3}$
$52=a\displaystyle \cdot\frac{-26}{-2}$
$52=13a\qquad/\times\frac{1}{13}$
$4=a$
The first term is 4.
