What is an AGP series (Arithmetico – Geometric Series) ?

It is a series where each term is a multiplication of a term in AP and a GP. The series $$a+(a+d) r+(a+2 d) r^2+\ldots+(a+(n-1) d) r^{n-1}$$ is known as the arithmetico-geometric series. Let’s denote the above sum by $$\begin{aligned}& S_n=a+(a+d) r+(a+2 d) r^2+\ldots+(a+(n-1) d) r^{n-1}\end{aligned}$$

What is an example of an AGP series?

$$1+3 \mathrm{x}+5 x^2+7 x^3+\ldots \ldots \ldots$$

Here 1, 3, 5,….. are in A.P. and $$1, \mathrm{x}, x^2, x^3\ldots \ldots \ldots$$ are in G.P.

AGP series formulas :-

(1) Sum of finite AGP Series formula:

$$S_n=\frac{a}{1-r}+\frac{d r}{(1-r)^2}-\frac{d r^n}{(1-r)^2}-\frac{(a+(n-1) d)r^n}{1-r}$$

(1) Sum of infinite AGP Series formula:

Impose the condition |r|<1, then $$\lim _{n \rightarrow \infty} r^n=0$$

So, sum to infinity of the AGP is$$S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}$$

NOTE:- Here you don’t need to learn the formula but you must know the process of solving questions of AGP.

In this article you will find all types of questions and related methods.

How to find solution of AGP related questions:

Let’s check out some questions of AGP series, where |x|<1.

Process:

Step 1: Multiplying equation by the common ratio.

Step 2: Write the final values one step forward so that you can easily subtract this from the given sum.

Q1. If |x|<1 then compute the sum of$$1+2 \mathrm{x}+3 x^2+4 x^3+\ldots \ldots \ldots$$

Solution:

$$\begin{aligned}& S=1+2 x+3 x^2+4 x^3+\ldots\end{aligned}$$

$$\begin{aligned}& x S=\quad x+2 x^2+3 x^3+\ldots\end{aligned}$$

“x” is the common ratio of the G.P. ( 1, x, x2 ,x3 ,……………………….)

Upon subtracting,$$\begin{aligned} & s(1-x)=1+x+x^2+x^3+\cdots\end{aligned}$$

$$\begin{aligned}s(1-x) & =\frac{1}{(1-x)} \end{aligned}$$

$$\begin{aligned}S_{\infty} & =\frac{1}{(1-x)^2}\end{aligned}$$

Q2. If |x|<1 then compute the sum of$$1+3 \mathrm{x}+6 x^2+10 x^3+\ldots \ldots \ldots$$

Solution:

$$\begin{aligned} & s=1+3 x+6 x^2+10 x^3+\cdots \end{aligned}$$

$$\begin{aligned} & sx=\quad x+3 x^2+6 x^3+\cdots\end{aligned}$$

“x” is the common ratio of the G.P. ( 1, x, x2 ,x3 ,……………………….)

Upon subtracting,$$\begin{aligned} &s(1-x)=1+2 x+3 x^2+4 x^3+\cdots\end{aligned}$$

Again multiplying by “x”

$$\begin{aligned} &s(1-x) x=\quad x+2 x^2+3 x^3+\cdots\end{aligned}$$

Upon subtracting,$$\begin{aligned} &s(1-x)(1-x)=1+x+x^2+x^3+\cdots\end{aligned}$$

$$\begin{gathered}\ s(1-x)(1-x)=\frac{1}{(1-x)} \end{gathered}$$

$$\begin{gathered}\ s_{\infty}=\frac{1}{(1-x)^{3}}\end{gathered}$$

Q3. Evaluate the following: [JEE (Main)-2019]

$$\begin{aligned}\sum_{k=1}^{20} k \frac{1}{2^{k}} \end{aligned}$$

Solution:

$$\begin{aligned}S=\sum_{k=1}^{20} k \frac{1}{2^{k}} \end{aligned}$$

$$\begin{aligned}S & =\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{20}{2^{20}}\end{aligned}$$

$$\begin{aligned}\quad\quad\quad\quad\quad\quad\frac{S}{2}=\quad\quad & \frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\cdots+\frac{19}{2^{20}}+\frac{20}{2^{21}}\end{aligned}$$

Upon subtracting,

$$\begin{aligned}S-\frac{S}{2} & =\left(\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{20}}\right)-\frac{20}{2^{21}} \end{aligned}$$

$$\begin{aligned}\frac{S}{2} & =1-\frac{1}{2^{20}}-\frac{20}{2^{21}} \end{aligned}$$

$$\begin{aligned}\frac{S}{2} & =1-\frac{1}{2^{20}}-\frac{10}{2^{20}} \end{aligned}$$

$$\begin{aligned}S & =2-\frac{1}{2^{19}}-\frac{10}{2^{19}} \end{aligned}$$

$$\begin{aligned}S & =2-\frac{11}{2^{19}} \end{aligned}$$

Q4. Find the value of k if

[JEE (Main)-2014]

$$\begin{aligned}& (10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\ldots+10(11)^{9}=k(10)^{9} \end{aligned}$$

Solution:

$$\begin{aligned}& k(10)^{9}=1(11)^{0}(10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\cdots+10(11)^{9}(10)^{0}\end{aligned}$$

$$\begin{aligned}& k(10)^{9} \frac{11}{10}=1(11)^{1}(10)^{8}+2(11)^{2}(10)^{7}+\cdots+9(11)^{9}(10)^{0}+\frac{10(11)^{10}}{10}\end{aligned}$$

$$\begin{aligned}& k(10)^{9}\left(1-\frac{11}{10}\right)=1(11)^{0}(10)^{9}+1(11)^{1}(10)^{8}+\cdots+1(11)^{9}(10)^{0}-11^{10}\end{aligned}$$

$$\begin{aligned}& k(10)^{9}\left(-\frac{1}{10}\right)=10^{9}\left(\frac{11^{10}}{10^{10}}-1\right) /\left(\frac{11}{10}-1\right)-11^{10}\end{aligned}$$

$$\begin{aligned}& -k(10)^{8}=10^{9}\left(\frac{11^{10}}{10^{10}}-1\right) /\left(\frac{1}{10}\right)-11^{10}\end{aligned}$$

$$\begin{aligned}& -k(10)^{8}=10^{10} \times \frac{11^{10}}{10^{10}}-10^{10}-11^{10}\end{aligned}$$

$$\begin{aligned}& -k(10)^{8}=11^{10}-10^{10}-11^{10}\end{aligned}$$

$$\begin{aligned}& -k(10)^{8}=-10^{10}\end{aligned}$$

$$\begin{aligned}& k=\frac{10^{10}}{10^{8}}=100\end{aligned}$$

AGP series questions

How to solve AGP series questions?

Sum of finite AGP formula

Sum of AGP series formula

Sum of infinite AGP series formula

1 thought on “What is an AGP series (Arithmetico – Geometric Series) ?”

Leave a Comment