Question on infinite geometric series. Problem- The sum of the first two terms of an infinite geometric series is 18. Also, each term of the series is seven times the sum of all the terms that follow. Find the first term and the common ratio of the series respectively.
  
My approach- Let $a+ar+ar^2+\dots$ be the series. Then,
$a+ar=18$ and, 
$a=7\frac{1}{1-r}-a$, 
solving I get $r = \frac{29}{43}$ but  Given answer is $a=16,r=1/8$.  where I'm doing wrong? 


The sum of the first two terms of an infinite geometric series is 18.

$$
S(a, r) = \sum_{k=0}^\infty a r^k = \frac{a}{1-r} \\
a + a r = 18 \quad (*)
$$

Also, each term of the series is seven times the sum of all the terms
  that follow.

$$
a r^n = 7 \sum_{k=n+1}^\infty a r^k \quad (**)
$$

Find the first term and the common ratio of the series respectively.

The first term is $a r^0 = a$ and the common ratio is
$$
\frac{a r^{n+1}}{a r^n} = r
$$
From $(*)$ we infer $a \ne 0$.
The instance $n=0$ of $(**)$ is
$$
a = 7 \sum_{k=1}^\infty a r^k
$$
so $a \ne 0$ implies $r \ne 0$.
We rewrite $(**)$ into
$$
\frac{1}{7}  r^n 
= \frac{1}{1-r} - \sum_{k=0}^n r^k
= \frac{1}{1-r} - \frac{1-r^{n+1}}{1-r}
= \frac{r^{n+1}}{1-r} \iff \\
\frac{1}{7} = \frac{r}{1-r} \iff \\
\frac{1}{7} - \frac{1}{7} r = r \iff \\
\frac{1}{7} = \frac{8}{7} r \iff \\
r = \frac{1}{8}
$$
We rewrite $(*)$ into
$$
18 = a (1+r) = a \frac{9}{8} \iff \\
a = 18 \frac{8}{9} = 16
$$
