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? 
 A: The second condition translates to :
$$a\times r^n=7\frac{a\times r^{n+1}}{1-r}\iff \frac{7r}{1-r}=1\iff r=\frac18$$
A: You are correct with the first equation $a+ar=18$. This implies that $a\neq 0$.  Applying the second condition to the first term, we get $$a=7(ar+ar^2+ar^3+\dots)$$ that is, we get
 $$a=7[-a+(a+ar+ar^2+ar^3+\dots)]$$ implying that
$$a=7\bigg[-a+\frac{a}{1-r}\bigg].$$ Because $a\neq 0$, we get
$$1=7\bigg[-1+\frac{1}{1-r}\bigg].$$ Solving for $r$, we get $r=\frac{1}{8}$. The value $a=16$ follows from the equation $a+ar=18$
A: 
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
$$
18 = a (1+r) = a \frac{9}{8} \iff \\
a = 18 \frac{8}{9} = 16
$$
