Arithmetico Geometric Series: Definition & Examples

What is an Arithmetico Geometric Series?

An arithmetico geometric series is a special type of series formed by multiplying the corresponding terms of an arithmetic progression (AP) and a geometric progression (GP). It combines the properties of both progressions, making it an important topic in sequences and series, higher algebra, mathematical analysis, and competitive examinations.

Arithmetico Geometric Series Meaning in Simple Words

In simple words, an arithmetico geometric series is a series in which each term is obtained by multiplying an arithmetic term and a geometric term.

For example, consider:

AP: $1, 2, 3, 4, 5,\ldots$

GP: $1, 2, 4, 8, 16,\ldots$

Multiplying corresponding terms gives:

$1,\ 4,\ 12,\ 32,\ 80,\ldots$

This new sequence is called an arithmetico geometric series.

Definition of Arithmetico Geometric Series

An arithmetico geometric series is a series whose terms are products of the corresponding terms of an arithmetic progression and a geometric progression.

The general form is:

$a+(a+d)r+(a+2d)r^2+(a+3d)r^3+\cdots$

where:

• $a$ = first term of the AP

• $d$ = common difference of the AP

• $r$ = common ratio of the GP

Since both AP and GP are involved, the series exhibits characteristics of arithmetic growth and geometric growth simultaneously.

Real-Life Examples of Arithmetico Geometric Series

Arithmetico geometric series appear in many practical situations involving increasing values combined with exponential growth or decay.

Example

Suppose a person invests an amount that increases by ₹100 every year while earning compound interest at a fixed rate.

Yearly investments:

₹100, ₹200, ₹300, ₹400, ...

Growth factors:

$1,\ r,\ r^2,\ r^3,\ldots$

The resulting value follows an arithmetico geometric series.

Why Arithmetico Geometric Series is Important

Arithmetico geometric series play an important role in advanced mathematics and practical modelling.

Importance of Arithmetico Geometric Series

• Combines concepts of AP and GP.

• Useful in higher algebra and calculus.

• Applied in economics and financial mathematics.

• Helps model real-world growth patterns.

• Important for sequence and series problems.

• Appears in mathematical competitions and entrance examinations.

• Used in engineering and computer science applications.

Structure of an Arithmetico Geometric Series

To understand an arithmetico geometric series, it is important to first understand its two building blocks: arithmetic progression and geometric progression.

Arithmetic Progression (AP) Component

An arithmetic progression is a sequence in which consecutive terms differ by a constant value called the common difference.

General AP

$a,\ a+d,\ a+2d,\ a+3d,\ldots$

where:

• $a$ = first term

• $d$ = common difference

Example

$2,\ 5,\ 8,\ 11,\ldots$

Here,

$d=3$

Geometric Progression (GP) Component

A geometric progression is a sequence in which consecutive terms are obtained by multiplying by a fixed number called the common ratio.

General GP

$1,\ r,\ r^2,\ r^3,\ldots$

where:

$r$ = common ratio

Example

$1,\ 2,\ 4,\ 8,\ 16,\ldots$

Here,

$r=2$

Formation of an Arithmetico Geometric Series

An arithmetico geometric series is formed by multiplying corresponding terms of an AP and a GP.

Example

AP: $1,\ 2,\ 3,\ 4,\ldots$

GP: $1,\ 2,\ 4,\ 8,\ldots$

Multiplying corresponding terms:

$1\times1,\ 2\times2,\ 3\times4,\ 4\times8,\ldots$

Resulting series:

$1,\ 4,\ 12,\ 32,\ldots$

This is an arithmetico geometric series.

General Form of an Arithmetico Geometric Series

The standard form of an arithmetico geometric series is:

$a+(a+d)r+(a+2d)r^2+(a+3d)r^3+\cdots$

Example

If: $a=2,\ d=3,\ r=2$

Then the series becomes:

$2+5(2)+8(2^2)+11(2^3)+\cdots$

$=2+10+32+88+\cdots$

Formula of Arithmetico Geometric Series

Several formulas are used to find terms and sums of arithmetico geometric series.

General Term Formula

The $n^{th}$ term of an arithmetico geometric sequence is:

$T_n=(a+(n-1)d)r^{n-1}$

where:

• $a$ = first term of AP

• $d$ = common difference

• $r$ = common ratio

• $n$ = term number

Example

If: $a=2,\ d=1,\ r=3$

Then: $T_4=(2+3)(3^3)$

$=5\times27$

$=135$

Sum of an Infinite Arithmetic-Geometric Series

When $|r|<1$, the infinite arithmetico-geometric series converges.

The sum is: $S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$

This formula is frequently used in advanced sequence and series problems.

Derivation of the Formula

Consider: $S=a+(a+d)r+(a+2d)r^2+\cdots$

Multiplying by $r$: $rS=ar+(a+d)r^2+(a+2d)r^3+\cdots$

Subtracting: $S-rS=a+dr+dr^2+dr^3+\cdots$

Therefore, $S(1-r)=a+d(r+r^2+r^3+\cdots)$

Using: $r+r^2+r^3+\cdots=\frac{r}{1-r}$

we obtain: $S(1-r)=a+\frac{dr}{1-r}$

Multiplying throughout by $(1-r)$:

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

This gives the infinite sum formula.

Important Parameters in the Formula

The following quantities are commonly used in arithmetico geometric series formulas.

Understanding these parameters makes formula application easier.

How to Find the Sum of an Arithmetico Geometric Series?

Finding the sum of an arithmetico geometric series involves identifying the AP and GP components and applying the appropriate formula.

Step-by-Step Method

Follow these steps:

Step 1

Identify the first term of the AP.

Step 2

Find the common difference $d$.

Step 3

Find the common ratio $r$.

Step 4

Determine whether the series is finite or infinite.

Step 5

Apply the required formula.

Method Using Formula

Consider the series: $1+2\left(\frac12\right)+3\left(\frac12\right)^2+\cdots$

Here: $a=1$

$d=1$

$r=\frac12$

Using: $S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$

Substituting values: $S_\infty=\frac{1}{1-\frac12}+\frac{1\cdot\frac12}{\left(1-\frac12\right)^2}$

$=2+2$

$=4$

Therefore, $S_\infty=4$

Infinite Series Cases

An infinite arithmetico geometric series has a finite sum only when: $|r|<1$

Examples

Convergent series: $r=\frac12,\ \frac13,\ -\frac14$

Divergent series: $r=2,\ 3,\ -5$

If the common ratio exceeds 1 in magnitude, the infinite sum does not exist.

Common Mistakes to Avoid

Students often make mistakes while solving arithmetico geometric series problems.

Common Errors

• Confusing AP with GP.

• Using the wrong common difference.

• Using the wrong common ratio.

• Applying infinite sum formulas when $|r|\ge1$.

• Forgetting to identify the correct first term.

• Mixing up term formulas and sum formulas.

Avoiding these mistakes can significantly improve accuracy in sequence and series questions for school examinations and competitive mathematics tests.

Learn more about Sequences and Series - Topics, Formula, Books, FAQs

Solved Example Based on Arithmetico-Geometric Progression

Example 1. Find the sum of the series $1+3+7+15+31+\cdots$ up to $n$ terms.

1. $2^{n+1}-n$
2. $2^{n+1}-n-2$
3. $2^n-n-2$
4. None of these

Hint: First find the general term of the series and then calculate the sum.

Solution:

Given,

$1+3+7+15+31+\cdots$

Observe that:

$1=2^1-1$

$3=2^2-1$

$7=2^3-1$

$15=2^4-1$

$31=2^5-1$

Therefore, the $n^{th}$ term is:

$T_n=2^n-1$

Now,

$S_n=\sum_{k=1}^{n}(2^k-1)$

$=\sum_{k=1}^{n}2^k-\sum_{k=1}^{n}1$

Using the GP sum formula:

$\sum_{k=1}^{n}2^k=\frac{2(2^n-1)}{2-1}$

$=2(2^n-1)$

$=2^{n+1}-2$

Also,

$\sum_{k=1}^{n}1=n$

Therefore,

$S_n=(2^{n+1}-2)-n$

$=2^{n+1}-n-2$

Correct Answer: $2^{n+1}-n-2$

Example 2. $S$ is the sum of the first 9 terms of $(x+ka)+(x^2+(k+2)a)+(x^3+(k+4)a)+\cdots$ where $a\neq0$ and $x\neq1$.

If $S=\frac{x^{10}-x+45a(x-1)}{x-1}$, find the value of $k$.

1. 3
2. 2
3. $-3$
4. $-5$

Hint: Separate the GP part and the AP part.

Solution:

Given,

$S=(x+ka)+(x^2+(k+2)a)+(x^3+(k+4)a)+\cdots$

Separating the two series,

$S=(x+x^2+x^3+\cdots+x^9)$

$+a[k+(k+2)+(k+4)+\cdots+(k+16)]$

The first series is a GP:

$x+x^2+x^3+\cdots+x^9$

$=\frac{x(x^9-1)}{x-1}$

The second series is an AP with 9 terms.

Sum of AP:

$=9k+(2+4+6+\cdots+16)$

Now,

$2+4+6+\cdots+16$

$=2(1+2+3+\cdots+8)$

$=2\left(\frac{8\times9}{2}\right)$

$=72$

Therefore,

$S=\frac{x(x^9-1)}{x-1}+a(9k+72)$

Comparing with:

$S=\frac{x^{10}-x+45a(x-1)}{x-1}$

we get,

$9k+72=45$

$9k=-27$

$k=-3$

Correct Answer: $-3$

Example 3. Which of the following is NOT an AGP?

1. $1,\ 2x,\ 3x^2,\ 4x^3$
2. $1,\ 1,\ \frac34,\ \frac12,\ \frac5{16},\ldots$
3. $3a,\ 5a^2,\ 7a^3$
4. $1,\ 3\cdot2^2,\ 4\cdot2^3,\ 5\cdot2^4$

Hint: Check whether each term can be expressed as the product of an AP term and a GP term.

Solution:

An AGP is formed by multiplying corresponding terms of an AP and a GP.

For Option 1:

$1,\ 2x,\ 3x^2,\ 4x^3$

This can be written as:

$(1,2,3,4)\times(1,x,x^2,x^3)$

Hence, it is an AGP.

For Option 2:

$1,\ 1,\ \frac34,\ \frac12,\ \frac5{16}$

This can be written as:

$\frac11,\ \frac22,\ \frac34,\ \frac48,\ \frac5{16}$

$=(1,2,3,4,5)\times\left(1,\frac12,\frac14,\frac18,\frac1{16}\right)$

Hence, it is an AGP.

For Option 3:

$3a,\ 5a^2,\ 7a^3$

This can be written as:

$(3,5,7)\times(a,a^2,a^3)$

Hence, it is an AGP.

For Option 4:

$1,\ 3\cdot2^2,\ 4\cdot2^3,\ 5\cdot2^4$

The arithmetic part should be:

$1,2,3,4,\ldots$

but here it becomes:

$1,3,4,5,\ldots$

which is not an AP.

Therefore, this is not an AGP.

Correct Answer: Option 4

Example 4. What is the sum of the first 10 terms of $1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+10\cdot2^{10}$ ?

1. $11\cdot2^{11}+2$
2. $9\cdot2^{11}+2$
3. $11\cdot2^{11}-2$
4. $9\cdot2^{11}-2$

Hint: Multiply the series by 2 and subtract.

Solution:

Let $S=1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+10\cdot2^{10}$

Multiplying by 2,

$2S=1\cdot2^2+2\cdot2^3+3\cdot2^4+\cdots+10\cdot2^{11}$

Subtracting,

$2S-S$

$=(1\cdot2^2-1\cdot2)$

$+(2\cdot2^3-2\cdot2^2)$

$+\cdots$

$+(10\cdot2^{11}-10\cdot2^{10})$

This simplifies to:

$S=10\cdot2^{11}-(2+2^2+2^3+\cdots+2^{10})$

Using GP sum:

$2+2^2+\cdots+2^{10}$

$=\frac{2(2^{10}-1)}{2-1}$

$=2^{11}-2$

Therefore,

$S=10\cdot2^{11}-(2^{11}-2)$

$=10\cdot2^{11}-2^{11}+2$

$=9\cdot2^{11}+2$

Correct Answer: $9\cdot2^{11}+2$

NCERT Resources

This section provides important NCERT study materials to help students strengthen their understanding of sequences and series. These resources are useful for concept building, board exam preparation, and competitive examinations.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

NCERT Class 11 Maths Chapter 9 Notes: Sequence and Series

NCERT Exemplar Class 11 Maths solutions Chapter 9 Sequence and Series

Related Topics to Sequence and Series

This section covers related mathematics topics that complement the study of sequences and series and help build a stronger foundation for algebra, progression-based problems, and competitive examinations.