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(1-r^(n+1))/(1-r)`

When
`|r|<1`, we have


`\displaystyle\lim_(n\to\infty)S_n=\frac a{1-r}`

as is the case here. Since
`S=\lim\limits_(n\to\infty)S_n=12` and
`r=\frac16` are given, we can find the first term
`a` immediately:


`12=\frac a{1-\frac16}\implies12=\frac65a\implies a=10`