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Hint: Any term in G.P. is equal to the sum of all the succeeding terms. We have:

$ \Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty $

The general term of G.P. can be written as:

$ \Rightarrow {T_n} = a{r^{n - 1}} .....(i)$

And according to the information given in the question, any term of the G.P. is equal to the sum of all the succeeding terms. From this we’ll get:

$ \Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty $

Substituting corresponding values in equation$(i)$, we’ll get:

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} + a{r^{n + 2}} + ......\infty ,$

$a$ is the first term of G.P. and its value is $1$ as per the information given in the question. So putting its value, we’ll get:

$

\Rightarrow {r^{n - 1}} = {r^n} + {r^{n + 1}} + {r^{n + 2}} + .....\infty , \\

\Rightarrow {r^{n - 1}} = {r^n}\left[ {1 + r + {r^2} + .....\infty } \right], \\

\Rightarrow \dfrac{1}{r} = \left[ {1 + r + {r^2} + .....\infty } \right] .....(ii) \\

$

Now, the terms on the right hand side of the above equation constitutes an infinite G.P. with $1$ as the first term and $r$ as the common ratio. And we know the formula for sum of terms of infinite G.P.:

$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$

So, on using this formula for equation $(ii)$,we’ll get:

$

\Rightarrow \dfrac{1}{r} = \dfrac{1}{{1 - r}}, \\

\Rightarrow 1 - r = r, \\

\Rightarrow 2r = 1, \\

\Rightarrow r = \dfrac{1}{2}. \\

$

Thus, the common ratio of the G.P. is $\dfrac{1}{2}$ and its first term is already given as $1$. So, we our infinite G.P.:

$ \Rightarrow 1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},.......\infty $

For finding sum of its terms, we will again apply${S_\infty } = \dfrac{a}{{1 - r}}$, we’ll get:

$

\Rightarrow {S_\infty } = \dfrac{1}{{1 - \dfrac{1}{2}}}, \\

\Rightarrow {S_\infty } = 2 \\

$

Therefore, the sum of infinite G.P. is $2$.

Note: If a G.P. consists of infinite terms, then we can only calculate the sum of its terms if it's common ratio is greater than $0$ and less than $1$$\left( {0 < r < 1} \right)$.Otherwise its sum will not be defined.

$ \Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty $

The general term of G.P. can be written as:

$ \Rightarrow {T_n} = a{r^{n - 1}} .....(i)$

And according to the information given in the question, any term of the G.P. is equal to the sum of all the succeeding terms. From this we’ll get:

$ \Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty $

Substituting corresponding values in equation$(i)$, we’ll get:

$ \Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} + a{r^{n + 2}} + ......\infty ,$

$a$ is the first term of G.P. and its value is $1$ as per the information given in the question. So putting its value, we’ll get:

$

\Rightarrow {r^{n - 1}} = {r^n} + {r^{n + 1}} + {r^{n + 2}} + .....\infty , \\

\Rightarrow {r^{n - 1}} = {r^n}\left[ {1 + r + {r^2} + .....\infty } \right], \\

\Rightarrow \dfrac{1}{r} = \left[ {1 + r + {r^2} + .....\infty } \right] .....(ii) \\

$

Now, the terms on the right hand side of the above equation constitutes an infinite G.P. with $1$ as the first term and $r$ as the common ratio. And we know the formula for sum of terms of infinite G.P.:

$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$

So, on using this formula for equation $(ii)$,we’ll get:

$

\Rightarrow \dfrac{1}{r} = \dfrac{1}{{1 - r}}, \\

\Rightarrow 1 - r = r, \\

\Rightarrow 2r = 1, \\

\Rightarrow r = \dfrac{1}{2}. \\

$

Thus, the common ratio of the G.P. is $\dfrac{1}{2}$ and its first term is already given as $1$. So, we our infinite G.P.:

$ \Rightarrow 1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},.......\infty $

For finding sum of its terms, we will again apply${S_\infty } = \dfrac{a}{{1 - r}}$, we’ll get:

$

\Rightarrow {S_\infty } = \dfrac{1}{{1 - \dfrac{1}{2}}}, \\

\Rightarrow {S_\infty } = 2 \\

$

Therefore, the sum of infinite G.P. is $2$.

Note: If a G.P. consists of infinite terms, then we can only calculate the sum of its terms if it's common ratio is greater than $0$ and less than $1$$\left( {0 < r < 1} \right)$.Otherwise its sum will not be defined.

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