Question

Show that in an infinite G.P. with common ratio r (|r| < 1), each term bears a constant ratio to the sum of all terms that follow it.

Answer 1

\[\text { Let us take a G . P . with terms }a_1 , a_2 , a_3 , a_4 , . . . \infty\text {  and common ratio r }\left( \left| r \right| < 1 \right) . \]

\[\text { Also, let us take the sum of all the terms following each term to be } S_1 , S_2 , S_3 , S_4 , . . . \]

\[\text { Now }, S_1 = \frac{a_2}{\left( 1 - r \right)} = \frac{ar}{\left( 1 - r \right)}, \]

\[ S_2 = \frac{a_3}{\left( 1 - r \right)} = \frac{a r^2}{\left( 1 - r \right)}, \]

\[ S_3 = \frac{a_4}{\left( 1 - r \right)} = \frac{a r^3}{\left( 1 - r \right)}, \]

\[ \Rightarrow \frac{a_1}{S_1} = \frac{a}{\frac{ar}{\left( 1 - r \right)}} = \frac{\left( 1 - r \right)}{r}, \]

\[\frac{a_2}{S_2} = \frac{ar}{\frac{a r^2}{\left( 1 - r \right)}} = \frac{\left( 1 - r \right)}{r}, \]

\[\frac{a_3}{S_3} = \frac{a r^2}{\frac{a r^3}{\left( 1 - r \right)}} = \frac{\left( 1 - r \right)}{r}, \]

\[\text { It is clearly seen that the ratio of each term to the sum of all the terms following it is constant . } \]