131k views
2 votes
Find the value of a1 for an infinite geometric series with S=12 and r=1/6

User Marson
by
6.8k points

1 Answer

2 votes

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
User Imranmadbar
by
7.5k points