Jambmaths question:
The sum to infinity of a geometric progression is $-\tfrac{1}{10}$and the first term is $-\tfrac{1}{8}$.Find the common ratio of the progression.
Option A:
$-\tfrac{1}{5}$
Option B:
$-\tfrac{1}{4}$
Option C:
$-\tfrac{1}{3}$
Option D:
$-\tfrac{1}{2}$
Jamb Maths Solution:
$\begin{align} & a=-\tfrac{1}{8} \\ & {{S}_{\infty }}=-\tfrac{1}{10} \\ & {{S}_{\infty }}=\frac{a}{1-r} \\ & -\frac{1}{10}=\frac{-\tfrac{1}{8}}{1-r} \\ & -\frac{1}{10}=-\frac{1}{8(1-r)} \\ & 8(1-r)=10 \\ & 8-8r=10 \\ & -8r=10-8 \\ & r=-\frac{2}{8} \\ & r=-\frac{1}{4} \\\end{align}$
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