# 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.

Last updated date: 19th Mar 2023

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Answer

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Hint: - Use the property, sum of infinite terms G.P as \[\dfrac{{{a_1}}}{{1 - r}}\]

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 \[{a_1} = 1\]

\[

\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} \\

\]

So the required 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{, }}.......................... \\

\]

So, this is the required answer.

Note: - In these types of questions the key concept is that always remember the sum of infinite terms G.P and the general series of infinite G.P, then according to given conditions calculate the value of common ratio, after getting this we can easily calculate the required infinite terms G.P series.

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 \[{a_1} = 1\]

\[

\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} \\

\]

So the required 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{, }}.......................... \\

\]

So, this is the required answer.

Note: - In these types of questions the key concept is that always remember the sum of infinite terms G.P and the general series of infinite G.P, then according to given conditions calculate the value of common ratio, after getting this we can easily calculate the required infinite terms G.P series.

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