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# Sum To n Terms Of a GP - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Sum of n-term of a GP is considered one of the most asked concept.

• 35 Questions around this concept.

## Solve by difficulty

The least positive integer n such that $\dpi{100} 1-\frac{2}{3}-\frac{2}{3^{2}}-............-\frac{2}{3^{n-1}}< \frac{1}{100}$  is

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

If each term of a geometric progression $\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots$ with $\mathrm{a}_1=\frac{1}{8}$ and $\mathrm{a}_2 \neq \mathrm{a}_1$, is the arithmetic mean of the next two terms and $\mathrm{S}_{\mathrm{n}}=\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{\mathrm{n}}$, then $\mathrm{S}_{20}-\mathrm{S}_{18}$ is equal to

## Concepts Covered - 3

Sum of n-term of a GP

Sum of n-term of a GP

Let Sn be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then

$\\S_n=a+ar+ar^2+...+ar^{n-2}+ar^{n-1}\;\;\;\;\;\;\;\;\;\;....(i)\\\mathrm{Multiply\;both\;side\;with\;\mathit{r}}\\rS_n=ar+ar^2+ar^3+...+ar^{n-1}+ar^{n}\;\;\;\;....(ii)\\\mathrm{Substract\;(ii)\;from\;(i)}\\S_n-rS_n=a-ar^{n}\\\\\mathrm{\Rightarrow S_n=\frac{a-ar^n}{1-r}=a\left ( \frac{1-r^n}{1-r} \right )}\\\\\mathrm{\Rightarrow S_n=a\left ( \frac{r^n-1}{r-1} \right )}$

The above formula does not hold for r = 1

For r = 1, each of the n terms is equal to a, and thus the sum of n terms of the G.P. is $S_n = na$.

Sum of an infinite GP

If a is the first term and r is the common ratio of a G.P. Then,

$\mathrm{S_n=a\left ( \frac{1-r^n}{1-r} \right )=\frac{a}{1-r}-\frac{ar^n}{1-r}}$

Let, - 1 < r < 1, i.e. | r | < 1, then

$\mathrm{\lim_{n\rightarrow \infty}r^n=0}$

[as this is infinite G.P., so n tends to infinity]

$\mathrm{S_{\infty}=\frac{a}{1-r}}$

$S_{\infty}$ is the sum to infinite terms of the G.P.

Note: If r ≥ 1, then the sum of an infinite G.P. tends to infinity.

Application of GP -Part 1

Application of GP

To write a non-terminating repeating number in p/q form:

Example:

The value of 0.358585858585..........

$\begin{array}{l}{0.3 \overline{5} \overline{8}=0.358585858 \ldots \ldots \text { to } \infty} \\ {\Rightarrow 0.3+0.058+0.00058+0.0000058+\ldots \ldots} \\ {\Rightarrow \frac{3}{10}+\frac{58}{10^{3}}+\frac{58}{10^{5}}+\frac{58}{10^{7}}+\ldots \ldots \ldots \ldots} \\\\ {\Rightarrow \frac{3}{10}+\frac{58}{10^{3}}\left(1+\frac{1}{10^{2}}+\frac{1}{10^{4}}+\ldots .\right)} \\\\ {\Rightarrow \frac{3}{10}+\frac{58}{10^{3}}\left(\frac{1}{1-\frac{1}{10^{2}}}\right)} \\\\ {\Rightarrow \frac{355}{990}}\end{array}$

Application of GP - Part 2

Application of GP

The sum of n-term of the series $a+a a+a a a+a a a a+\ldots \ldots \ldots ., \forall a \in \mathbb{N}, 1 \leq a \leq 9$ is

$\frac{a}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-n\right]$

The sum of n-term of the series $0 . a+0 . a a+0 . a a a+0 . a a a a+\ldots \ldots ., \quad \forall a \in \mathbb{N}, \quad 1 \leq a \leq 9$  is

$\frac{a}{9}\left\{n-\frac{1}{9}\left[1-\left(\frac{1}{10}\right)^{n}\right]\right\}$

## Study it with Videos

Sum of n-term of a GP
Application of GP -Part 1
Application of GP - Part 2

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## Books

### Reference Books

#### Sum of n-term of a GP

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

Page No. : 5.14

Line : 1

#### Application of GP -Part 1

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

Page No. : 5.15

Line : 1