Question

In a geometric series, if the sum of the first 3 terms is -3/8 and the sum to infinity is -1/3, what is the common ratio?

Answer 1

sn =a(1-r^n)/(1-r)
=sinf(1-r^n) since sinf =a/(1-r)
(-3/8)=(-1/3)(1-r^3)
(1-r^3)=9/8
r^3=-1/8
r=-1/2

Source: http://www.mathskey.com/question2answer/

Answer 2

\[S_{\infty}=\frac{1}{1-r}\]

Answer 3

plug in what you know \(S_{\infty}=-\large{\frac{1}{3}}\) and solve for r (the common ratio)

Answer 4

oops, should be \[S_\infty=\frac{ a_n }{ 1-r }\]

Answer 5

can use this and the fact that \(S_3 = -\large{\frac{3}{8}}\) to find r

Answer 6

Let the first term be a and common ratio r.
\[a+ar+ar^2=-\frac{ 3 }{ 8 },a \left( 1+r+r^2 \right)=-\frac{ 3 }{ 8 }\]
\[S _{\infty}=\frac{ a }{ 1-r },-\frac{ 1 }{ 3 }=\frac{ a }{ 1-r },a=-\frac{ 1-r }{ 3 }\]
\[-\frac{ 1-r }{ 3 }\left( 1+r+r^2 \right)=-\frac{ 3 }{ 8 }\]
\[1-r^3=-\frac{ 3 }{ 8 }\times- 3=\frac{ 9 }{ 8 },-r^3=\frac{ 9 }{ 8 }-1=\frac{ 9-8 }{ 8 }=\frac{ 1 }{ 8 }\]
\[r^3=-\frac{ 1 }{ 8 }=\left( \frac{ -1 }{ 2 } \right)^3,r=-\frac{ 1 }{ 2 }\]