Question

The first term of an infinite G.P is 1 and any term is equal to the sum of all the succeeding terms. Find the series
$
  (a){\text{ 1,2,4,8}}........ \\
  {\text{(b) 1,}}\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},......... \\
  (c){\text{ 1,}}\dfrac{1}{4},\dfrac{1}{8},....... \\
  (d){\text{ 1,}}\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}.......... \\
$

Answer 1

Hint – In this question consider an infinite G.P of the form \[{a_1},{\text{ }}{a_1}r,{\text{ }}{a_1}{r^2}{\text{, }}{a_1}{r^3}{\text{, }}........................\infty \], then use the formula that sum of infinite terms of a G.P, that is \[{a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........................\infty = \dfrac{{{a_1}}}{{1 - r}}\], try and compute the value of r, using the value of first term of the G.P. This will help get the right series.
Complete step-by-step answer:
It is given that the first term of an infinite G.P is 1.
\[ \Rightarrow {a_1} = 1\]
Now, we know the sum of infinite G.P\[\left( {{S_\infty }} \right) = \dfrac{{{a_1}}}{{1 - r}}\], (where r is the common ratio)
Let the infinite G.P series is
\[{a_1},{\text{ }}{a_1}r,{\text{ }}{a_1}{r^2}{\text{, }}{a_1}{r^3}{\text{, }}........................\infty \]
Therefore the sum of this series is
\[{S_\infty } = {a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........................\infty = \dfrac{{{a_1}}}{{1 - r}}................\left( 1 \right)\]
Now according to question it is given that any term is equal to the sum of succeeding terms
\[ \Rightarrow {a_1} = {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........................\infty \]
Now add both sides by \[{a_1}\]
\[ \Rightarrow {a_1} + {a_1} = {a_1} + {a_1}r + {a_1}{r^2} + {a_1}{r^3} + ........................\infty \]
From equation (1)
\[ \Rightarrow {\text{2}}{a_1} = \dfrac{{{a_1}}}{{1 - r}}\]
Now it is given that \[{a_1} = 1\]
\[\begin{gathered}
   \Rightarrow {\text{2}} \times {\text{1}} = \dfrac{1}{{1 - r}} \\
   \Rightarrow 1 - r = \dfrac{1}{2} \Rightarrow r = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\
\end{gathered} \]
So the required series is
\[{a_1},{\text{ }}{a_1}r,{\text{ }}{a_1}{r^2}{\text{, }}{a_1}{r^3}{\text{, }}........................\infty \]
\[ = 1,{\text{ }}\dfrac{1}{2}{\text{, }}{\left( {\dfrac{1}{2}} \right)^2}{\text{, }}{\left( {\dfrac{1}{2}} \right)^3}{\text{, }}{\left( {\dfrac{1}{2}} \right)^4}{\text{, }}..........................\infty \]
\[ = 1,{\text{ }}\dfrac{1}{2}{\text{, }}\left( {\dfrac{1}{4}} \right){\text{, }}\left( {\dfrac{1}{8}} \right){\text{, }}\left( {\dfrac{1}{{16}}} \right){\text{, }}..........................\infty \]
So, this is the required answer.
Hence option (D) is the correct answer.

Note – A series is said to be in G.P if and only if the common ratio that is the ratio of consecutive terms of the series remains constant throughout the series, if we have a look at the general series that we have considered that is \[{a_1},{\text{ }}{a_1}r,{\text{ }}{a_1}{r^2}{\text{, }}{a_1}{r^3}{\text{, }}........................\infty \], then the common ratio is $\dfrac{{{a_1}r}}{{{a_1}}} = r{\text{ or }}\dfrac{{{a_1}{r^2}}}{{{a_1}r}} = r$, clearly it remains constant throughout, thus our series prediction of this to be in G.P is absolutely right.