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Question
Find the sum of the following geometric progression:
(a2 − b2), (a − b), \[\left( \frac{a - b}{a + b} \right)\] to n terms;
Solution
Here, a = a2 − b2 and r = \[\frac{1}{a + b}\]
\[\therefore S_n = a\left( \frac{1 - r^n}{1 - r} \right) \]
\[ = \left( a^2 - b^2 \right) \left( \frac{1 - \left( \frac{1}{a + b} \right)^n}{1 - \left( \frac{1}{a + b} \right)} \right) \]
\[ = \left( a^2 - b^2 \right)\left( \frac{\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right)^n} \right)}{\frac{\left( a + b \right) - 1}{a + b}} \right)\]
\[ \Rightarrow S_n = \frac{\left( a + b \right)\left( a - b \right)}{\left( a + b \right)^{n - 1}}\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right) - 1} \right)\]
\[ = \frac{\left( a - b \right)}{\left( a + b \right)^{n - 2}}\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right) - 1} \right)\]
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