Question

Find the value of a1 for an infinite geometric series with S=12 and r=1/6

Answer 1

$S_n=a(1+r+\cdots+r^{n-1}+r^n)$
$rS_n=a(r+r^2+\cdots+r^n+r^{n+1})$
$(1-r)S_n=a(1-r^{n+1})$
$S_n=a\dfrac{1-r^{n+1}}{1-r}$

When $|r|$, we have

$\displaystyle\lim_{n\to\infty}S_n=\dfrac a{1-r}$

as is the case here. Since $S=\lim\limits_{n\to\infty}S_n=12$ and $r=\dfrac16$ are given, we can find the first term $a$ immediately:

$12=\dfrac a{1-\frac16}\implies12=\dfrac65a\implies a=10$