The sum to infinity of a geometric progression is \(-\frac{1}{10}\) and the first term is \(-\frac{1}{8}\). Find the common ratio of the progression.
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Correct Answer: Option B
Explanation:
Sr = \(\frac{a}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8} \times \frac{1}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8(1 - r)}\)
\(-\frac{1}{10}\) = \(\frac{1}{8 - 8r}\)
cross multiply...
-1(8 - 8r) = -10
-8 + 8r = -10
8r = -2
r = -1/4
Sr = \(\frac{a}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8} \times \frac{1}{1 - r}\)
\(-\frac{1}{10}\) = \(\frac{1}{8(1 - r)}\)
\(-\frac{1}{10}\) = \(\frac{1}{8 - 8r}\)
cross multiply...
-1(8 - 8r) = -10
-8 + 8r = -10
8r = -2
r = -1/4