Sum of First n Terms of GP Formula

Geometric Progression

A geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The 'constant factor' is called the common ratio and is denoted by ' $r$ '. $r$ is also a non-zero number.

The first term of a G.P. is usually denoted by $a$.

If $\mathrm{\mathit{a_1,a_2,a_3.....a_{n-1},a_n}}$is in geometric progression

$\mathrm{then,\;\mathit{r=\frac{a_2}{a_1}=\frac{a_3}{a_2}=....=\frac{a_n}{a_{n-1}}}}$

Eg,

• $2,6,18,54, \ldots .(a=2, r=3)$
• $4,2,1,1 / 2,1 / 4, \ldots .(a=4, r=1 / 2)$
• $-5,5,-5,5, \ldots \ldots . .(a=-5, r=-1)$

General Term of a GP

If ' $a$ ' is the first term and ' $r$ ' is the common ratio, then

$a_1=a=a r^{1-1}\left(1^{\text {st }}\right.$ term $)$

$a_2=a r=a r^{2-1}\left(2^{\text {nd }}\right.$ term $)$

$a_3=a r^2=a r^{3-1} \quad\left(3^{\text {rd }}\right.$ term $)$

$a_n=a r^{n-1}\left(\mathrm{n}^{\mathrm{th}}\right.$ term $)$

So, the general term or $n^{\text {th }}$ term of a geometric progression is $a_n=ar^{n-1}$

Geometric Mean

If three terms are in G.P., then the middle term is called the Geometric Mean (G.M.) of the other two numbers. So if $\mathrm{a}, \mathrm{b}$, and $
\mathrm{c} \text { are in G.P., then } \mathrm{b} \text { is } \mathrm{GM} \text { of } \mathrm{a} \text { and } \mathrm{c} \text {, }
$

If $a_1, a_2, a_3, \ldots ., a_n$ are n positive numbers, then the Geometric Mean of these numbers is given by $G=\sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot \ldots . . \cdot a_n}$

If $a$ and $b$ are two numbers and $G$ is the $G M$ of $a$ and $b$. Then, $a, G, \underline{b}$ are in geometric progression.

Hence, $G=\sqrt{a \cdot b}$

$
\text { Insertion of n-Geometric Mean Between a and b }
$

Let $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}$ be n geometric mean between two numbers a and b . Then, $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}, b$ is an G.P. Clearly, this G.P. contains $\mathrm{n}+2$ terms.
now, $\mathrm{b}=(\mathrm{n}+2)^{\text {th }}$ term $=\mathrm{ar}^{\mathrm{n}+2-1}$
$\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$
[where, $\mathrm{r}=$ common ratio]

$
\begin{aligned}
& \therefore \mathrm{G}_1=\mathrm{ar}, \mathrm{G}_2=\mathrm{ar}^2, \mathrm{G}_3=\mathrm{ar}^3, \ldots ., \mathrm{Ga}_{\mathrm{n}}=\mathrm{ar}^{\mathrm{n}} \\
& \Rightarrow \mathrm{G}_1=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{1}{\mathrm{n}+1}}, \mathrm{G}_2=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{2}{\mathrm{n}+1}}, \mathrm{G}_3=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{3}{\mathrm{n}+1}} \ldots \ldots \\
& \mathrm{G}_{\mathrm{n}}=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{n}}{\mathrm{a}+1}}
\end{aligned}
$

The sum of $ \text { n } $ term of a GP

Let $S_n$ 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 + \cdots + ar^{n-2} + ar^{n-1} \quad \text{(i)}
$

Multiply both sides by \( r \):

$
rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n-1} + ar^n \quad \text{(ii)}
$

Subtract (ii) from (i):

$
S_n - rS_n = a - ar^n
$

Thus,

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

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

The above formula does not hold for $r=1$
For $\mathrm{r}=1$, each of the n terms is equal to $a$, and thus the sum of $n$ terms of the G.P. is $S_n = na$.

$
\text { So, The sum of } n \text { terms of GP with first term 'a' and common ratio ' } r \text { ' is given by }
$

The sum of an infinite GP

Consider an infinite GP $a, a r, a r^2, a r^3, a r^4 \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

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

The last term is not known.

Then,

$
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]

The sum of an infinite term of GP is given by

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

$S_{\infty}$ is the sum of infinite terms of the G.P.Note: If $r \geq 1$, then the sum of an infinite G.P. tends to infinity.

Application of GP

• Using Geometric Progression, we can convert non-terminating numbers into a fraction.

To write a non-terminating repeating number in $
\text { p/q form: }
$

Example:

In this example, we will convert the non-terminating number into a fraction.

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} $

• The sum of the n-term of the series

$a+a a+a a a+a a a a+\ldots \ldots \ldots ., \forall a \in \mathbb{N}, 1 \leq a \leq 9$

With the use of GP, we can find the sum of the n term of the above GP

The sum of the $ n $ term of GP is given by

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

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

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

Recommend Video Based on Sum of n Terms of GP

Solved Examples Based on the Sum of n Terms of GP

Example 1: Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in \mathbb{N}$, be two G.P.s with common ratios $\mathrm{r}_1$ and $\mathrm{r}_2$ respectively such that $\mathrm{a}_1=\mathrm{b}_1=4$ and $\mathrm{r}_1<\mathrm{r}_2$. Let $c_{\mathrm{k}}=\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}, \mathrm{k} \in \mathbb{N}$. If $\mathrm{c}_2=5$ and $c_3=\frac{13}{4} \sum_{\text {then } \mathrm{k}=1}^{\infty} \mathrm{c}_{\mathrm{k}}-\left(12 \mathrm{a}_6+8 \mathrm{~b}_4\right)$ is equal to

[JEE MAINS 2023]

Solution:

$
\begin{aligned}
&\begin{array}{ll}
\mathrm{a}_1=4 & \text { GP } 4,4 \mathrm{r}_1, 4 \mathrm{r}_1^2- \\
\mathrm{b}_1=4 & \text { GP } 4,4 \mathrm{r}_2, 4 \mathrm{r}_2^2- \\
\mathrm{C}_2=\mathrm{a}_2+\mathrm{b}_2 \quad \mathrm{C}_3=\mathrm{a}_3+\mathrm{b}_3 \\
5=4 \mathrm{r}_1+4 \mathrm{r}_2 & \frac{13}{4}=4 \mathrm{r}_1^2+4 \mathrm{r}_2^2 \\
\frac{5}{4}=\mathrm{r}_1+\mathrm{r}_2 \ldots(1) \quad \mathrm{r}_1^2+\mathrm{r}_2^2=\frac{13}{16} \ldots \\
\frac{25}{16}=r_1^2+r_2^2+2 r_1 r_2 \\
\frac{25}{16}=\frac{13}{16}+2 r_1 r_2 \\
\Rightarrow r_1 r_2=\frac{12}{16 \times 2}=\frac{3}{8}
\end{array}\\
&\text { (2) }
\end{aligned}
$

$
\begin{aligned}
& \text { Now } r_1+\frac{3}{8 r_1}=\frac{5}{4} \\
& 8 r_1^2+3=10 r_1 \\
& \Rightarrow 8 r_1^2-10 r_1+3=0 \\
& r_1=\frac{3}{4}, r_1=\frac{1}{2} \\
& r_2=\frac{1}{2} r_2=\frac{3}{4} \\
& \because r_1<r_2 \\
& r_1=\frac{1}{2} \\
& \therefore \quad r_2=\frac{3}{4}
\end{aligned}
$

Now $C_k=a_k+b_k$

$
\begin{aligned}
& \sum_{\mathrm{k}=1}^{\infty} \mathrm{C}_{\mathrm{k}}=\frac{4}{1-\mathrm{r}_1}+\frac{4}{1-\mathrm{r}_2} \\
& =\frac{4}{1-\frac{1}{2}}+\frac{4}{1-\frac{3}{4}} \\
& =8+16=24 \\
& \sum_{\mathrm{k}=1}^{\infty} \mathrm{C}_{\mathrm{k}}-\left(12 \mathrm{a}_6+8 \mathrm{~b}_4\right) \Rightarrow 24-\left\{12 \times 4\left(\frac{1}{2}\right)^5+8 \times 4\left(\frac{3}{4}\right)^3\right\} \\
& =24-\left(12 \times \frac{1}{8}+8 \times \frac{27}{16}\right) \\
& =24-\left\{\frac{3}{2}+\frac{27}{2}\right\} \\
& =24-15 \\
& =9
\end{aligned}
$

Hence, the required answer is 9.

Example 2:

The sum of 20 terms of the series $2.2^2-3^2+2.4^2-5^2+2.6^2-\ldots \ldots \ldots$ is equal to [JEE MAINS 2023]

Solution:

$\begin{aligned} & \left(2^2-3^2+4^2-5^2+20 \text { terms }\right)+\left(2^2+4^2+\ldots+10 \text { terms }\right) \\ & -(2+3+4+5+\ldots+11)+4\left[1+2^2+\ldots 10^2\right] \\ & -\left[\frac{21 \times 22}{2}-1\right]+4 \times \frac{10 \times 11 \times 21}{6} \\ & =1-231+14 \times 11 \times 10 \\ & =1540+1-231 \\ & =1310\end{aligned}$

Hence, the required answer is 1310.

Example 3:

Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2, G_3$ be three geometric means of two distinct positive numbers. Then $G_1^4+G_3^4+G_3^4+G_3^2 G_3^2$ is equal to

[JEE MAINS 2023]

Solution

$ \begin{aligned}
& \quad a, \mathrm{~A}_1, \mathrm{~A}_2, \text { bareinA.P. } \\
& d=\frac{b-a}{3}: A_1=a+\frac{b-a}{3}=\frac{2 a+b}{3} \\
& A_2=\frac{a+2 b}{3} \\
& A_1+A_2=a+b \\
& a, G_1, G_2, G_{s,} b \text { are in } G \cdot P \\
& r=\left(\frac{b}{a}\right)^{\frac{1}{4}} \\
& G_1=\left(a^3 b\right)^{\frac{1}{4}} \\
& G_2=\left(a^2 b^2\right)^{\frac{1}{4}} \\
& G_5=\left(a b^3\right)^{\frac{1}{4}}
\end{aligned} $

Example 4:

The 4th term of GP is 500 and its common ratio is $\frac{1}{m}, m \in \mathrm{N}_{\text {. }}$ Let $S_{\mathrm{n}}$ denote the sum of the first n terms of this GP. If $\mathrm{S}_6>\mathrm{S}_5+1$ and $\mathrm{S}_7<\mathrm{S}_6+\frac{1}{2}$, then the number of possible values of m is

[JEE MAINS 2023]

Solution

$
\mathrm{T}_4=500 \Rightarrow \mathrm{a}\left(\frac{1}{\mathrm{~m}}\right)^3=500 \Rightarrow \mathrm{a}=500 \mathrm{~m}^3
$
Now $\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}=\mathrm{a}\left(\frac{1-\mathrm{r}^{\mathrm{n}}}{1-\mathrm{r}}\right)-\mathrm{a}\left(\frac{1-\mathrm{r}^{\mathrm{n}-1}}{1-\mathrm{r}}\right)$

$
\begin{aligned}
& =\frac{a}{1-r}\left[r^{n-1}(1-r)\right] \\
& =a r^{n-1} \\
& =500 \mathrm{~m}^3\left(\frac{1}{m}\right)^{n-1} \\
& S_n-S_{n-1}=500 \mathrm{~m}^{4-n}
\end{aligned}
$

Now $S_6-S_s>1 \Rightarrow 500 \mathrm{~m}^{-2}>1 \ldots$
$\& S_7-S_6<\frac{1}{2} \Rightarrow 500 \mathrm{~m}^{-3}<\frac{1}{2} \ldots$
$\left.\begin{array}{lc}\text { from(1) } & \mathrm{m}^2<500 \\ \text { from(2) } & \mathrm{m}^3>1000\end{array}\right] 10<\mathrm{m} \leq 22$
The number of possible values of $m$ is $=12$

Example 5:

For $\mathrm{k} \in \mathrm{N}$, if the sum of the series $1+\frac{4}{\mathrm{k}}+\frac{8}{\mathrm{k}^2}+\frac{13}{\mathrm{k}^3}+\frac{19}{\mathrm{k}^4}+\ldots \ldots$ is 10 , then the value of k is:

[JEE MAINS 2023]

Solution

$\begin{aligned} & 10=1+\frac{4}{\mathrm{k}}+\frac{8}{\mathrm{k}^2}+\frac{13}{\mathrm{k}^3}+\frac{19}{\mathrm{k}^4}+\ldots . . \text { upto } \infty \\ & 9=\frac{4}{\mathrm{k}}+\frac{8}{\mathrm{k}^2}+\frac{13}{\mathrm{k}^3}+\frac{19}{\mathrm{k}^4}+\ldots . \text { upto } \infty \\ & \frac{9}{\mathrm{k}}=\frac{4}{\mathrm{k}^2}+\frac{8}{\mathrm{k}^3}+\frac{13}{\mathrm{k}^4}+\ldots . . \text { upto } \infty \\ & \mathrm{S}=9\left(1-\frac{1}{\mathrm{k}}\right)=\frac{4}{\mathrm{k}}+\frac{4}{\mathrm{k}^2}+\frac{5}{\mathrm{k}^3}+\frac{6}{\mathrm{k}^4} \ldots . . \text { upto } \infty \\ & \frac{\mathrm{S}}{\mathrm{k}}=\frac{4}{\mathrm{k}^2}+\frac{4}{\mathrm{k}^3}+\frac{5}{\mathrm{k}^4}+\ldots \ldots \text { upto } \infty \\ & \left(1-\frac{1}{\mathrm{k}}\right) \mathrm{S}=\frac{4}{\mathrm{k}}+\frac{1}{\mathrm{k}^3}+\frac{1}{\mathrm{k}^4}+\frac{1}{\mathrm{k}^5}+\ldots . \infty \\ & \left(1-\frac{1}{\mathrm{k}}\right)^2=\frac{4}{\mathrm{k}}+\frac{\frac{1}{\mathrm{k}^3}}{\left(1-\frac{1}{\mathrm{k}}\right)} \\ & 9(\mathrm{k}-1)^3=4 \mathrm{k}(\mathrm{k}-1)+1 \\ & k=2\end{aligned}$

Hence, the answer is 2.