
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 sum of the infinite series.
Answer
602.7k+ views
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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

