Question

Find the sum of the terms of an infinite decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is equal to 32/81.

Answer 1

\[\text{ Let r be the common ratio of the given G . P } . \]

\[ \therefore a = 4\]

\[\text { Sum of the geometric ifinite series: } \]

\[ S_\infty = 4 + 4r + 4 r^2 + . . . \infty \]

\[\text { Now, } S_\infty = \frac{4}{1 - r} . . . . . . . \left( i \right)\]

\[\text { The difference between the third and fifth term is } \frac{32}{81} . \]

\[ a_3 - a_5 = \frac{32}{81}\]

\[ \Rightarrow 4 r^2 - 4 r^4 = \frac{32}{81}\]

\[ \Rightarrow 4\left( r^2 - r^4 \right) = \frac{32}{81}\]

\[ \Rightarrow 81 r^4 - 81 r^2 + 8 = 0 . . . . . . . \left( ii \right)\]

\[\text { Now, let } r^2 = y\]

\[\text { Let us put this in } \left( ii \right) . \]

\[ \therefore 81 r^4 - 81 r^2 + 8 = 0\]

\[ \Rightarrow 81 y^2 - 81y + 8 = 0\]

\[ \Rightarrow 81 y^2 - 72y - 9y + 8 = 0\]

\[ \Rightarrow 9y\left( 9y - 1 \right) - 8\left( 9y - 1 \right) = 0\]

\[ \Rightarrow \left( 9y - 8 \right)\left( 9y - 1 \right)\]

\[ \Rightarrow y = \frac{1}{9}, \frac{8}{9}\]

\[\text { Putting y } = r^2 ,\text {  we get } r = \frac{1}{3} \text { and } \frac{2\sqrt{2}}{3}\]

\[\text { Substituting r } = \frac{1}{3} \text { and }\frac{2\sqrt{2}}{3} \text { in } \left( i \right): \]

\[ S_\infty = \frac{4}{1 - \frac{1}{3}} = \frac{12}{2} = 6\]

\[\text { Similarly }, S_\infty = \frac{4}{1 - \frac{2\sqrt{2}}{3}} = \frac{12}{3 - 2\sqrt{2}}\]

\[ \therefore S_\infty = 6, \frac{12}{3 - 2\sqrt{2}}\]