Arithmetic Geometric Series: Definition, General Terms, Examples

Arithmetic Geometric Series: Think about a restaurant where a square table fits \(4\) people. When two square tables are put together. then \(6\) people are seated. Put \(3\) square tables together, and then \(8\) people are seated and so on. This is an example of an arithmetic sequence as the numbers \(4,6,8….\) are increasing by adding \(2\) to the previous term. Now, if we add all these terms of an arithmetic sequence. then we obtain an arithmetic series. The name suggests that it has something to do with arithmetic series and geometric series. Let us see more about the Arithmetic Geometric Series in the following article.

What is a Sequence?

A set of numbers arranged in a definite order according to a rule (or rules) is called a sequence. Each number of the set is called a term of the sequence. A sequence is called finite or infinite according to the number of terms in it is finite or infinite. The different terms of a sequence are usually denoted by \(a_{1}, a_{2}, a_{3}, \ldots\) or \({T_1},{T_2},{T_3},…..\) The subscript (always a natural number) denotes the position of the term in the sequence. The number occurring at the \(n^{t h}\) place of a sequence i.e \(a_{n}\) is called the general term of the sequence.

Example: \(3,5,7,9, \ldots, 21\) and \(1,-\frac{1}{2}, \frac{1}{4},-\frac{1}{8}, \ldots\)

What is a Series?

If \(T_{1}, T_{2}, T_{3}, T_{4}, \ldots, T_{n}, \ldots\) is a sequence, then the expression \(T_{1}+T_{2}+T_{3}+T_{4}+\cdots+T_{n}+\cdots\) is a series.
There are different types of series based on the rule followed by the terms. Two of the commonly used types of series are given below.

• Arithmetic series
• Geometric series

Let us recall the arithmetic and geometric series.

Arithmetic Geometric Sequence

The sequence whose each term is formed by multiplying the corresponding terms of an A.P. and G.P. is called arithmetic-geometric sequence.

The general or standard form of such a sequence is given by

\(a,(a+d) r_{,}(a+2 d) r^{2}, \ldots\)

Here,

A.P. \(=a, a+d, a+2 d, \ldots\)

G.P \(=1, r, r^{2}, \ldots\)

This sequence is also called arithmetic geometric progression (A.G.P.).

Arithmetic- Geometric Series

Let \(a, a+d, a+2 d, \ldots, a+(n-1) d\) be an arithmetic sequence, and \(1, r, r^{2}, \ldots, r^{n-1}\) be a geometric sequence.

Then, the term-by-term multiplication of an arithmetic sequence with the corresponding terms of a geometric sequence gives us a new sequence known as an arithmetic-geometric sequence.

Thus, the terms of an arithmetic-geometric sequence are given as follows

\(a,(a+d) r,(a+2 d) r^{2}, \ldots[a+(n-1) d] r^{n-1}\)

When the terms of an arithmetic-geometric sequence are added, an arithmetic-geometric series is obtained.

Thus, the arithmetic-geometric series of \(n\) term is

\(a+(a+d) r+(a+2 d) r^{2}+\cdots+(a+(n-1) d) r^{n-1}\)

Example: The series \(1+3 x+6 x^{2}+9 x^{3}+12 x^{4}+15 x^{5}+18 x^{6}+\cdots\) is an arithmetic-geometric series obtained by multiplying the terms of an arithmetic sequence \(1,3,6,9, \ldots\) with the corresponding terms of a geometric sequence \(1, x, x^{2}, x^{3}, \ldots\)

\({n^{th}}\) Term of Arithmetic Geometric Series

Consider the standard form of the arithmetic-geometric series

\(a,\,(a+d) r,\,(a+2 d) r^{2}, \ldots\)

Since the \(n^{t h}\) term of an A.P \( a,\,(a+d),\,(a+2 d), \ldots\) is given by

\(a_{n}=a+(n-1) d\)

And the \(n^{t h}\) term of the G.P \(1, r, r^{2}, \ldots\) is given by

\(r^{n-1}\)

Therefore, the \(n^{t h}\) term of the arithmetic-geometric series

\(a_{n}=[a+(n-1) d] r^{n-1}\)

Remarks:

1. When we substitute \(r=1\) in the above arithmetic-geometric series, we get \(a + (a + d) + (a + 2d) + \cdots + (a + (n – 1)d)\). which is an arithmetic series of \(n\) terms.
2. When we substitute \(d=0\) in the above arithmetic-geometric series, we get \(a + ar + a{r^2} + \cdots + a{r^{n – 1}}\) which is a geometric series of \(n\) terms.

Sum to n-Terms of an Arithmetic Geometric Series

Let \(a_{,}(a+d) r,(a+2 d) r^{2}, \ldots\) be the given sequence and let \(S_{n}\) be the sum \(n\) terms. Then \(S_{n}=a+(a+d) r+(a+2 d) r^{2}+\cdots+(a+(\mathrm{n}-2) d) r^{n-2}+(a+(\mathrm{n}-1) d) r^{n-1}\) …..(i)

Multiplying equation (i) by the common ratio \(r\) we have

\(r S_{n}=a r+(a+d) r^{2}+(a+2 d) r^{3}+\cdots \ldots+(a+(n-2) d) r^{n-1}+(a+(n-1) d) r^{n}\) ……(ii)

Subtracting equation (ii) from equation (i) we have

\(S_{n}(1-r)=\left(a+d r+d r^{2}+\cdots\right.\) to \(n\) terms \({)} -(a+(\mathrm{n}-1) d) r^{n}\)

\(=a+\left(d r+d r^{2}+\cdots\right.\) to \((\mathrm{n}-1)\) terms \({)}-(a+(\mathrm{n}-1) d) r^{n}\)

\(=a+\frac{d r\left(1-r^{n-1}\right)}{(1-r)}-(a+(\mathrm{n}-1) d) r^{n}\)

\(\therefore S_{n}=\frac{a}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^{2}}-\frac{(a+(\mathrm{n}-1) d) r^{n}}{1-r}\)

When \(r \neq 1\)

Note that remembering this formula may be difficult. So in the problems, we follow the whole procedure instead of using the formula.

Remark: When \(r=1\)

The arithmetic-geometric series, we get is \(a+(a+d)+(a+2 d)+\cdots+(a+(n-1) d)\) which is an A.P

And, the sum of \(n\) terms of an A.P. is given by

\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)

Steps to Find the Sum of an Arithmetic Geometric Series

Follow the algorithm to find the sum of an arithmetic geometric series:

Step 1: Let the given series equal \(S_{n}\) and consider it equation(i)

Step 2: Multiply the equation (i) by the common ratio of the given geometric progression involved in the given series.

Step 3: Put the equation obtained in step \(2\) be equation (ii).

Step 4: Either subtract equation (ii) from equation (i) or add both the equations so that the new equation should involve a geometric progression.

Step 5: Simplify the equation obtained in step \(4\) by applying the formula for sum of geometric series.

Sum to Infinity of an Arithmetic Geometric Series

We have already seen that sum of \(n\) terms of arithmetic- geometric series is

\(S_{n}=\frac{a}{1-r}+\frac{d r\left(1-r^{n-1}\right)}{(1-r)^{2}}-\frac{(a+(\mathrm{n}-1) d) r^{n}}{1-r}\) provided \(r \neq 1\)

If \(|r|<1\), as \(n \rightarrow \infty, r^{n-1}, r^{n}, n r^{n} \rightarrow 0\)

If \(S\) denotes the limiting sum, then

\(S=\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}\)

Solved Examples – Arithmetic Geometric Series

Q.1. Find the sum of infinite series \(1-3 x+5 x^{2}-7 x^{3}+\cdots\) when \(|x|<1\)
Ans: Consider the given series
\(1-3 x+5 x^{2}-7 x^{3}+\cdots\) when \(|x|<1\)
We can see that this series is arithmetic geometric series which is formed by multiplying the corresponding terms of A.P i.e., \(1,3,5,7, \ldots\) and the terms of G.P i.e., \(1,-x, x^{2},-x^{3}, \ldots\)
Let \(S=1-3 x+5 x^{2}-7 x^{3}+\cdots \) upto \(\infty\)…..(i)
Multiply both sides of (i) by the common ratio of G.P. i.e, \(-x\)
Thus, we have
\(x \mathrm{~S}=x-3 x^{2}+5 x^{3}-\cdots\) upto \(\infty\) ……(ii)
Adding (i) and (ii), we have
\((1+x) S=1-2 x+2 x^{2}-2 x^{3}+\cdots \infty\)
\(=1-2 x\left(1-x+x^{2}-x^{3}+\cdots\right.\) to \(\left.\infty\right)\)
\(=1-\frac{2 x}{1+x}=\frac{1-x}{1+x}\) \({[} \because \) Sum of infinite G.P. i.e…, \(a+a r+a r^{2}+\cdots\) upto \(\infty\) is given by \(\left.\frac{a}{1-r}\right]\)
Hence, the sum of the given series \(=\frac{1-x}{(1+x)^{2}}\)

Q.2. Find the sum to infinity of the series \(1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\cdots\)
Ans: Given series is \(1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\cdots\)
This series is an Arithmetic-geometric series formed by the combination of term of Arithmetic Progression i.e., \(1,\,2,\,3,\,4….\) and Geometric Progression i.e., \(1, \frac{1}{3}, \frac{1}{3^{2}}, \cdots\) And, common ratio of G.P \( = \frac{1}{3} < 1\)
Let \(S=1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\cdots\) to \(\infty\)……(i)
Multiplying each side of the above equation by the common ratio \(\frac{1}{3}\), we have
\(\frac{1}{3} S=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\cdots\) to \(\infty\) ……(ii)
Subtracting equation (ii) from equation (i), we have
\(\left(1-\frac{1}{3}\right) S=1+\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\cdots\)
\(\Rightarrow \frac{2}{3} S=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\)
Hence, \(S=\frac{9}{4}\) is the required sum.

Q.3. Find the sum to \(n\) terms of the series \(\frac{2}{3}+\frac{5}{3^{2}}+\frac{8}{3^{3}}+\cdots\)
Ans: Given series is
\(\frac{2}{3}+\frac{5}{3^{2}}+\frac{8}{3^{3}}+\cdots\)
This series is an Arithmetic-Geometric series, as it is formed by multiplying the corresponding terms of A.P. \(2,\,5,\,8……\) and G.P \(\frac{1}{3}, \frac{1}{3^{2}}, \frac{1}{3^{3}}, \ldots\)Thus, the nth term of the given series is
\([2+(n-1) 3]\left[\frac{1}{3}\left(\frac{1}{3}\right)^{n-1}\right]=(3 n-1)\left(\frac{1}{3}\right)^{n}=\frac{3 n-1}{3^{n}}\)
Let \(S_{n}=\frac{2}{3}+\frac{5}{3^{2}}+\frac{8}{3^{3}}+\cdots+\frac{3 n-4}{3^{n-1}}+\frac{3 n-1}{3^{n}}\) …….(i)
Multiplying both sides of the above equation by \(\frac{1}{3}\), we have
\(\frac{1}{3} S_{n}=\frac{2}{3^{2}}+\frac{5}{3^{3}}+\cdots \cdots \cdots+\frac{3 n-4}{3^{n}}+\frac{3 n-1}{3^{n+1}}\) ……(ii)
Subtracting equation (ii) from equation (i), we have
\(\frac{2}{3} S_{n}=\frac{2}{3}+\left(\frac{3}{3^{2}}+\frac{3}{3^{3}}+\cdots+\frac{3}{3^{n}}\right)-\frac{3 n-1}{3^{n+1}}\)
\(=\frac{2}{3}+\frac{3}{3^{2}}\left(\frac{1-\left(\frac{1}{3}\right)^{n-1}}{1-\frac{1}{3}}\right)-\frac{3 n-1}{3^{n+1}}\)
\(\therefore S_{n}=1+\frac{3}{4}\left(1-\frac{1}{3^{n-1}}\right)-\frac{3 n-1}{2 \cdot 3^{n}}\) is the required sum.

Q.4. If the sum to infinity of the series
\(3+(3+d) \frac{1}{4}+(3+2 d) \cdot \frac{1}{4^{2}}+\cdots\) is \(3 \frac{1}{3}\), find \(d\)
Ans: Let \(S=3+(3+d) \cdot \frac{1}{4}+(3+2 d) \cdot \frac{1}{4^{2}}+\cdots\) to \(\infty\) ….(i)
Multiplying both sides of (i) by \(\frac{1}{4}\) we have
\(\frac{1}{4} S=\frac{3}{4}+(3+d) \frac{1}{4^{2}}+(3+2 d) \frac{1}{4^{3}}+\cdots+\infty\) …..(ii)
Subtracting (ii) from (i), we have
\(\frac{3}{4} S=3+\frac{1}{4} d+\frac{1}{4^{2}} d+\cdots\)
\(=3+\frac{\frac{d}{4}}{1-\frac{1}{4}}=3+\frac{d}{3}\)
It is given that \(S=3 \frac{1}{3}=\frac{10}{3}\)
Thus, \(\frac{3}{4} S=3+\frac{d}{3}\)
\(\Rightarrow \frac{3}{4} \times \frac{10}{3}=3+\frac{d}{3}\)
\(\Rightarrow \frac{5}{2}=3+\frac{d}{3}\)
\(\Rightarrow \frac{d}{3}=\frac{5}{2}-3\)
Hence, \(d=-\frac{3}{2}\) is the required value.

Q.5. Find the sum to \(n\) terms of the series \(2+5 x+8 x^{2}+11 x^{3}+\cdots,|x|<1\)
Ans: Given series is \(2+5 x+8 x^{2}+11 x^{3}+\cdots,|x|<1\)
We  can see that the above series is formed by multiplying corresponding terms of A.P. \(2,5,8,11….\) and G.P \(1, x, x^{2}, x^{3}, \ldots\)
Hence, \(n^{t h}\) term of given arithmetic-geometric series is
\([2+(n-1) 3] \cdot x^{n-1}=(3 n-1) x^{n-1}\)
Let \(S_{n}=2+5 x+8 x^{2}+11 x^{3}+\cdots+(3 n-4) x^{n-2}+(3 n-1) x^{n-1}\)…(i)
Multiplying both sides of (i) by \(x\) we have
\(x S_{n}=2 x+5 x^{2}+8 x^{3}+\cdots+(3 n-4) x^{n-1}+(3 n-1) x^{n}\)…..(ii)
\(\Rightarrow(1-x) \mathrm{S}_{n}=2+\left(3 x+3 x^{2}+3 x^{3}+\cdots+3 x^{n-1}\right)-(3 n-1) x^{n}\)
\(=2+3 x\left(\frac{1-x^{n-1}}{1-x}\right)-(3 n-1) x^{n}\)
\(\therefore S_{n}=\frac{2}{1-x}+\frac{3 x\left(1-x^{n-1}\right)}{(1-x)^{2}}-\frac{(3 n-1) x^{n}}{1-x}\) is the required sum.

Summary

If the difference between any term and its preceding term is always the same, then the sequence is called an arithmetic progression. If in a sequence of non-zero numbers, the ratio of a term to the term before it is always a constant quantity, then a sequence is called a geometric progression (abbreviated as G.P.).. And, the series which is formed by multiplying the corresponding terms of an A.P. and G.P. is called arithmetic-geometric sequence. We have discussed the method to find the sum of n terms and the sum of infinite terms of any called arithmetic-geometric series. We have also provided few solved examples based on this concept for more clarity.

Frequently Asked Questions (FAQs)

Q.1. How do you find the sum of an arithmetic geometric series?
Ans: Follow the following steps to find the sum of an arithmetic geometric series:
Step 1: Let the given series equal \(S_{n}\) and consider it equation (i)
Step 2: Multiply the equation (i) by the common ratio of the given geometric progression involved in the given series.
Step 3: Put equation obtain in step \(2\) be equation (ii).
Step 4: Either subtract equation (ii) from equation (i) or add both the equation so that the ne equation should involve a geometric progression.
Step 5: Simplify the equation obtained in step \(4\) by applying the formula for sum of geometric series.

Q.2. What is arithmetic series and geometric series?
Ans: The sum of terms in an arithmetic sequence is called an arithmetic series. Similarly, the sum of terms in a geometric sequence is called a geometric series.

Q.3. What is a series and a sequence? 
Ans: A sequence is an ordered arrangement of numbers according to some rule. The terms of a sequence are generally denoted by \(a_{1}, a_{2}, a_{3}, \ldots\) or \(T_{1}, T_{2}, T_{3}, \ldots\) The sum of terms in a sequence is known as series.
Thus, if \(a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots\) is a sequence then \(a_{1}+a_{2}+a_{3}+\cdots+a_{n}+\cdots\) is a series.

Q.4. How do you know if a sequence is arithmetic or geometric?
Ans: The following pointers are to be noted while determining a sequence is arithmetic or geometric:

• If the common difference of all terms in a sequence is the same, the sequence is arithmetic.
• If the common ratio of all terms in a sequence is the same, it is geometric sequence.

Q.5. What is the formula for arithmetic geometric progression? 
Ans: The progression whose term is formed by multiplying the corresponding terms of an A.P. and G.P. is called arithmetic geometric sequence.
The general or standard form of such a progression is given by:
\(a,(a+d) r,(a+2 d) r^{2}, \ldots\)

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