Question

Find the sum of the following geometric progression:

(a² − b²), (a − b), \[\left( \frac{a - b}{a + b} \right)\] to n terms;

Answer 1

Here, a = a² − b² and r = \[\frac{1}{a + b}\]

\[\therefore S_n = a\left( \frac{1 - r^n}{1 - r} \right) \]

\[ = \left( a^2 - b^2 \right) \left( \frac{1 - \left( \frac{1}{a + b} \right)^n}{1 - \left( \frac{1}{a + b} \right)} \right) \]

\[ = \left( a^2 - b^2 \right)\left( \frac{\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right)^n} \right)}{\frac{\left( a + b \right) - 1}{a + b}} \right)\]

\[ \Rightarrow S_n = \frac{\left( a + b \right)\left( a - b \right)}{\left( a + b \right)^{n - 1}}\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right) - 1} \right)\]

\[ = \frac{\left( a - b \right)}{\left( a + b \right)^{n - 2}}\left( \frac{\left( a + b \right)^n - 1}{\left( a + b \right) - 1} \right)\]