Sum of an Infinite Arithmetic Geometric Series

What is an Arithmetic Geometric Progression (AGP)?

An Arithmetic-Geometric Progression (AGP) is a special type of series where each term is formed by multiplying corresponding terms of an Arithmetic Progression (AP) and a Geometric Progression (GP). Unlike pure AP or GP, AGP displays characteristics of both linear and exponential growth, making it more complex and interesting in the domain of sequences and series.

Definition of AGP in Sequence and Series

An AGP is a series where each term can be expressed as:

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

Here:

• $a$ is the first term of the arithmetic part,

• $d$ is the common difference of the arithmetic part,

• $r$ is the common ratio of the geometric part,

• $n$ is the term number.

This blend gives the series its unique structure. A common example is:

$S = a \cdot r^0 + (a + d) \cdot r^1 + (a + 2d) \cdot r^2 + \cdots$

This form is frequently used to derive the sum of AGP formula and plays a key role in finding the sum of infinite terms of AGP.

Difference Between AP, GP, and AGP

Understanding the differences between these sequences is crucial:

• AP (Arithmetic Progression): Terms increase or decrease by a constant difference.
  Example: $2, 5, 8, 11, \dots$ (Common difference $d = 3$)

• GP (Geometric Progression): Terms change by a constant ratio.
  Example: $3, 6, 12, 24, \dots$ (Common ratio $r = 2$)

• AGP (Arithmetic Geometric Progression): Each term is a product of an AP term and a GP term.
  Example: $2 \cdot 1 + 5 \cdot \frac{1}{2} + 8 \cdot \frac{1}{4} + 11 \cdot \frac{1}{8} + \cdots$

While AP and GP have direct summation formulas, the sum of AGP series formula requires special techniques such as recursive subtraction or term-wise manipulation.

Common Patterns and Examples in AGP

Typical AGP questions involve recognising patterns like:

• $S = 1 + 2r + 3r^2 + 4r^3 + \cdots$

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

These patterns help in applying the correct AGP infinite sum formula or finite sum of AGP formula, depending on the convergence condition ($|r| < 1$ for infinite series). One popular example is:

$S = \sum_{n=1}^{\infty} n \cdot r^{n-1} = \frac{1}{(1 - r)^2}$, where $|r| < 1$.

Such series often appear in questions asking for the sum of an infinite AGP, particularly in competitive exams and algebraic problem-solving.

General Formula for the Sum of AGP

The sum of an Arithmetic-Geometric Progression (AGP) is not as direct as in AP or GP. Because each term is a combination of both linear and exponential growth, the summation requires a strategic approach. The AGP sum formula differs based on whether the series is finite or infinite. Understanding the sum of the AGP series formula is essential when dealing with hybrid sequences in algebra.

AGP Sum Formula for a Finite Number of Terms

For a finite AGP of the form:
$S_n = \sum_{k=0}^{n-1} (a + kd) \cdot r^k$
where:

• $a$ is the first term of the arithmetic part,

• $d$ is the common difference,

• $r$ is the common ratio of the geometric part,

• $n$ is the number of terms,

The sum of the AGP formula is derived using algebraic manipulation and recursive techniques, which we'll cover below.

Step-by-Step Derivation of AGP Sum Formula

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

We split this into two parts:

• One sum involving $a$: $a(1 + r + r^2 + \cdots + r^{n-1}) = a \cdot \frac{1 - r^n}{1 - r}$

• Another involving $d$: $d(r + 2r^2 + 3r^3 + \cdots + (n - 1)r^{n-1})$

This second part is handled by using the identity:
$\sum_{k=1}^{n-1} kr^k = \frac{r(1 - nr^{n-1} + (n - 1)r^n)}{(1 - r)^2}$

Combining both gives the complete AGP sum formula for finite terms.

Use of the Recursive Substitution Method in AGP

A common method for summing AGP involves recursive substitution.
Let: $S = a + (a + d)r + (a + 2d)r^2 + \cdots$

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

Subtract the two equations to cancel terms and reduce:
$S - rS = a + dr + dr^2 + dr^3 + \cdots$

The remaining terms form a geometric series, which can now be summed using the standard GP sum formula.

Sum of an Infinite AGP Series

When $|r| < 1$, the AGP becomes convergent, and its sum tends toward a finite value. This is where the sum of the infinite AGP formula becomes useful in higher mathematics and exams.

Conditions for Infinite AGP to Converge

For the sum of infinite AGP to exist, the geometric ratio $r$ must satisfy: $|r| < 1$

This ensures that higher-order terms approach zero, allowing the series to converge. If this condition is not met, the series diverges and the sum does not exist.

Derivation of the Infinite AGP Sum Formula

Let: $S = a + (a + d)r + (a + 2d)r^2 + (a + 3d)r^3 + \cdots$

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

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

The RHS becomes a GP:
$d \cdot \left( \frac{r}{1 - r} \right)$ and
$a \cdot \left( \frac{1}{1 - r} \right)$

Thus, the sum of an infinite number of AGP terms is:
$S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}$

The complete formula is: $S = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}$, where $|r| < 1$

Solved Examples Based on Arithmetico-Geometric Series

Example 1: If the sum of the series $\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^2}-\frac{1}{2 \cdot 3}+\frac{1}{3^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^2 \cdot 3}+\frac{1}{2 \cdot 3^2}-\frac{1}{3^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^3 \cdot 3}+\frac{1}{2^2 \cdot 3^2}-\frac{1}{2 \cdot 3^3}+\frac{1}{3^4}\right)+\cdots$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha + 3\beta$ is equal to [JEE MAINS 2023]

Solution: The sum of the series represents an Arithmetico-Geometric Progression,
$P = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2} \right) + \left( \frac{1}{2^3} + \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^2} \right) + \cdots$

$P \left( \frac{1}{2} + \frac{1}{3} \right) = \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \left( \frac{1}{2^3} + \frac{1}{3^3} \right) + \left( \frac{1}{2^4} - \frac{1}{3^4} \right) + \cdots$

$\frac{5P}{6} = \frac{\frac{1}{6}}{1 - \frac{1}{2}} - \frac{\frac{1}{8}}{1 + \frac{1}{3}}$

$\frac{5P}{6} = \frac{1}{2} - \frac{1}{12} = \frac{5}{12}$

$\therefore P = \frac{6}{5} \cdot \frac{5}{12} = \frac{1}{2}$

$\therefore P = \frac{\alpha}{\beta}, \quad \alpha = 1, \beta = 2$

$\alpha + 3\beta = 1 + 3 \cdot 2 = 7$

Hence, the required answer is 7.

Example 2: Suppose $\mathrm{a}_1, \mathrm{a}_2, 2, \mathrm{a}_3, \mathrm{a}_4$ be in an arithemetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric 49 progression is $\frac{49}{2}$, then $a_4$ is equal to [JEE MAINS 2023]

Solution: Given,
Common ratio $=\mathrm{r}=2$

The sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$
$\text{Terms: } \frac{(a - 2d)}{4}, \frac{(a - d)}{2}, a, 2(a + d), 4(a + 2d)$

$a = 2$

$\left( \frac{1}{4} + \frac{1}{2} + 1 + 6 \right) \cdot 2 + (-1 + 2 + 8)d = \frac{49}{2}$

$2 \left( \frac{3}{4} + 7 \right) + 9d = \frac{49}{2}$

$2 \cdot \frac{31}{4} + 9d = \frac{49}{2}$

$\frac{62}{4} + 9d = \frac{49}{2}$

$9d = \frac{49}{2} - \frac{62}{4} = \frac{98 - 62}{4} = \frac{36}{4} = 9$

$\Rightarrow d = 1$

$\Rightarrow a_4 = 4(a + 2d) = 4(2 + 2) = 16$

Hence, the required answer is 16.

Example 3: The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\frac{51}{6^5}+\frac{70}{6^6}+\ldots$ is equal to:

1) $\frac{425}{216}$

2) $\frac{429}{216}$

3) $\frac{288}{125}$

4) $\frac{280}{125}$

Solution: $\text { Let } S=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\cdots$

$\therefore \quad \frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\cdots$

Subtracting these

$\frac{5 \mathrm{~S}}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\cdots$

$\frac{5 \mathrm{~S}}{6} \cdot \frac{1}{6}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\cdots$

Subtracting

$\frac{5 \mathrm{~S}}{6}\left(1-\frac{1}{6}\right)=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\cdots$

$\Rightarrow \frac{25 \mathrm{~S}}{36}=1+\frac{3}{6} \frac{1}{\left(1-\frac{1}{6}\right)}=1+\frac{3}{5}$

$\Rightarrow \frac{25 \mathrm{~S}}{36}=\frac{8}{5}$

$\Rightarrow \mathrm{~S}=\frac{288}{125}$

Hence, the answer is option 3.

Example 4: The sum $1+2 \cdot 3+3 \cdot 3^2+\ldots \ldots . .+10 \cdot 3^9$ is equal to :

1) $\frac{2 \cdot 3^{12}+10}{4}$

2) $\frac{19 \cdot 3^{10}+1}{4}$

3) $5 \cdot 3^{10}-2$

4) $\frac{9 \cdot 3^{10}+1}{2}$

Solution: $\mathrm{S}_{\mathrm{n}}=1+2 \cdot 3^1+3 \cdot 3^2+4 \cdot 3^3+\ldots .+10 \cdot 3^9$

$3 \mathrm{~S}_{\mathrm{n}}=1 \cdot 3^1+2 \cdot 3^2+3 \cdot 3^2+\ldots . .+10 \cdot 3^{10}$

$-2 \mathrm{~S}_{\mathrm{n}}=1+3+3^2+\ldots . .+3^9-10 \cdot 3^{10}$

$\quad=\frac{1\left(3^{10}-1\right)}{3-1}-10 \cdot 3^{10}$

$\quad=\frac{1}{2} \cdot 3^{10}-\frac{1}{2}-3^{10} \cdot 10$

$-2 \mathrm{~S}_{\mathrm{n}}=\frac{-1}{2}-\frac{19}{2} \cdot 3^{10}$

$\mathrm{~S}_{\mathrm{n}}=\frac{19}{4} \cdot 3^{10}+\frac{1}{4}$

Hence, the answer is the option (2).

Example 5: Let $\mathrm{S}=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+\ldots$ Then 4 S is equal to [JEE MAINS 2022]
Solution: The given series represents an Arithmetico-Geometric Progression,
$S = 2 + \frac{6}{7} + \frac{12}{7^2} + \frac{20}{7^3} + \frac{30}{7^4} + \cdots$

$\frac{S}{7} = \frac{2}{7} + \frac{6}{7^2} + \frac{12}{7^3} + \frac{20}{7^4} + \cdots$

$\text{Subtracting: } \quad S - \frac{S}{7} = 2 + \frac{4}{7} + \frac{6}{7^2} + \frac{8}{7^3} + \frac{10}{7^4} + \cdots$

$\Rightarrow \frac{6}{7} S = 2 + \frac{4}{7} + \frac{6}{7^2} + \frac{8}{7^3} + \cdots$

$\Rightarrow \frac{6}{7} S = 2 \left(1 + \frac{2}{7} + \frac{3}{7^2} + \frac{4}{7^3} + \cdots \right)$

$\text{Now let: } T = 1 + \frac{2}{7} + \frac{3}{7^2} + \frac{4}{7^3} + \cdots$

$\text{This is a standard AGP: } T = \sum_{n=1}^{\infty} \frac{n}{7^{n-1}}$

$T = \frac{1}{(1 - \frac{1}{7})^2} = \frac{49}{36}$

$\Rightarrow \frac{6}{7} S = 2 \cdot \frac{49}{36}$

$\Rightarrow \frac{6}{7} S = \frac{98}{36}$

$\Rightarrow S = \frac{98 \cdot 7}{36 \cdot 6} = \frac{686}{216} = \frac{343}{108}$

$\Rightarrow 4S = \left(\frac{7}{3}\right)^3$

Hence, the required answer is $\left(\frac{7}{3}\right)^3$

List of Topics related to the Sum of Infinite AGP

To fully understand the sum of infinite AGP, it's helpful to explore key related concepts like Arithmetic Progression, Geometric Progression, and Harmonic Progression. This section covers essential definitions, formulas, and examples from each to strengthen your grasp on sequences and series.

NCERT Resources

Access important NCERT resources for Class 11 Chapter 9: Sequence and Series, including detailed notes, step-by-step solutions, and curated exemplar problems to build a strong foundation for AGP and other series-based topics.

NCERT Notes Class 11 Maths Chapter 9 Sequence and Series

NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series

NCERT Exemplar Solutions Class 11 Maths Chapter 9 Sequence and Series

Practice Questions based on the Sum of an Infinite AGP

Strengthen your understanding of the sum of an infinite AGP with targeted MCQs and practice problems. This section includes formula-based, conceptual, and exam-style questions covering AGP, special series, and related techniques like the method of differences and AM-GM inequality.

Sum Of An Infinite Arithmetic-Geometric Series Practice Question MCQ

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