Interactive Demo: Visualize AGP
See how arithmetic and geometric progressions combine to form AGP patterns.
Real-Life Hook: The Compound Interest with Installments
You invest ₹1000 in year 1, ₹2000 in year 2, ₹3000 in year 3, and so on (arithmetic increase). But each investment grows at 10% per year (geometric growth). How much total wealth do you have after many years?
This creates a sequence where:
- Year 1 contribution grows: $1000 \times (1.1)^{n-1}$
- Year 2 contribution grows: $2000 \times (1.1)^{n-2}$
- Year 3 contribution grows: $3000 \times (1.1)^{n-3}$
This is an Arithmetico-Geometric Progression - where arithmetic and geometric patterns combine!
Understanding AGP is crucial for financial mathematics, probability distributions, and complex JEE problems!
What is an AGP?
An Arithmetico-Geometric Progression (AGP) is a sequence where each term is the product of corresponding terms from an AP and a GP.
General Form:
If AP is: $a, a+d, a+2d, a+3d, \ldots$
And GP is: $1, r, r^2, r^3, \ldots$
Then AGP is: $a, (a+d)r, (a+2d)r^2, (a+3d)r^3, \ldots$
Standard Form:
$$\boxed{a, (a+d)r, (a+2d)r^2, (a+3d)r^3, \ldots, [a+(n-1)d]r^{n-1}}$$Example: $1, 4x, 7x^2, 10x^3, 13x^4, \ldots$
Here: AP is $1, 4, 7, 10, 13, \ldots$ (with $a=1, d=3$)
GP is $1, x, x^2, x^3, x^4, \ldots$ (with ratio $r=x$)
nth Term of AGP
The $n$th term of AGP is:
$$\boxed{T_n = [a + (n-1)d] \cdot r^{n-1}}$$Structure:
- AP part: $a + (n-1)d$
- GP part: $r^{n-1}$
- AGP term: Their product!
Memory Trick: “AP Times GP Power” → $[a+(n-1)d] \times r^{n-1}$
Sum of n Terms of AGP
This is the KEY FORMULA for JEE!
For AGP: $a, (a+d)r, (a+2d)r^2, \ldots$
$$\boxed{S_n = \frac{a}{1-r} + \frac{dr[1-r^{n-1}]}{(1-r)^2} - \frac{[a+(n-1)d]r^n}{1-r}}$$Alternative Form (cleaner for memory):
When $r \neq 1$:
$$\boxed{S_n = \frac{a - [a+(n-1)d]r^n}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2}}$$When $r = 1$: AGP becomes AP!
$$S_n = na + \frac{n(n-1)d}{2}$$Derivation Method: Multiply sum by $r$ and subtract (see below)
Derivation of Sum Formula
Method: Multiply by r and Subtract
$$S_n = a + (a+d)r + (a+2d)r^2 + \cdots + [a+(n-1)d]r^{n-1}$$Multiply by $r$:
$$rS_n = ar + (a+d)r^2 + (a+2d)r^3 + \cdots + [a+(n-1)d]r^n$$Subtract:
$$S_n - rS_n = a + dr + dr^2 + dr^3 + \cdots + dr^{n-1} - [a+(n-1)d]r^n$$ $$S_n(1-r) = a + dr(1 + r + r^2 + \cdots + r^{n-2}) - [a+(n-1)d]r^n$$The part $(1 + r + r^2 + \cdots + r^{n-2})$ is a GP sum with $(n-1)$ terms:
$$= \frac{1-r^{n-1}}{1-r}$$So:
$$S_n(1-r) = a + dr \cdot \frac{1-r^{n-1}}{1-r} - [a+(n-1)d]r^n$$ $$S_n = \frac{a}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2} - \frac{[a+(n-1)d]r^n}{1-r}$$Sum of Infinite AGP
When $|r| < 1$, the infinite sum exists:
$$\boxed{S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}}$$Derivation:
As $n \to \infty$ and $|r| < 1$:
- $r^{n-1} \to 0$
- $r^n \to 0$
From finite sum formula:
$$S_\infty = \frac{a}{1-r} + \frac{dr(1-0)}{(1-r)^2} - \frac{[a+\infty \cdot d] \cdot 0}{1-r}$$ $$= \frac{a}{1-r} + \frac{dr}{(1-r)^2}$$Simplified:
$$\boxed{S_\infty = \frac{a(1-r) + dr}{(1-r)^2} = \frac{a - ar + dr}{(1-r)^2} = \frac{a + r(d-a)}{(1-r)^2}}$$Memory Trick: “A plus D-A times R, all over (1-R) squared”
Special Cases
Case 1: $a = 1, d = 1$
AGP: $1, 2r, 3r^2, 4r^3, \ldots$
$$\boxed{S_n = 1 + 2r + 3r^2 + 4r^3 + \cdots + nr^{n-1}}$$Infinite sum when $|r| < 1$:
$$\boxed{S_\infty = \frac{1}{(1-r)^2}}$$This is a super important formula for JEE!
Case 2: $a = d$
AGP: $a, 2ar, 3ar^2, 4ar^3, \ldots$
$$S_\infty = \frac{a}{(1-r)^2}$$Common Mistakes & How to Avoid Them
❌ Mistake 1: Wrong nth Term
Wrong: $T_n = ar^{n-1} + (n-1)d$ ✗ (addition instead of multiplication!)
Correct: $T_n = [a + (n-1)d] \cdot r^{n-1}$ ✓
Key: It’s a product, not sum!
❌ Mistake 2: Infinite Sum Convergence
Wrong: $S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ for any $r$ ✗
Correct: Only when $|r| < 1$ ✓
Example: For $r = 2$, the series $1 + 4 + 12 + 32 + \cdots$ diverges!
❌ Mistake 3: Formula Confusion
Problem: Find sum of $1 + 3x + 5x^2 + 7x^3 + \cdots$ to $n$ terms
Wrong: Using GP formula ✗
Correct: This is AGP with $a = 1, d = 2, r = x$
Use AGP sum formula! ✓
❌ Mistake 4: Sign in $(1-r)$
Problem: Sum $1 - 2x + 3x^2 - 4x^3 + \cdots$
Wrong: Using $r = x$ ✗
Correct: Here $r = -x$ (negative!)
$$S_\infty = \frac{1}{[1-(-x)]^2} = \frac{1}{(1+x)^2}$$Problem-Solving Strategies
Strategy 1: Identify AP and GP Components
Problem: Find the 5th term of AGP: $2, 6x, 12x^2, 20x^3, \ldots$
Solution:
Identify pattern:
- Coefficients: $2, 6, 12, 20, \ldots$
- Check differences: $6-2=4$, $12-6=6$, $20-12=8$ (not AP!)
- Check if product pattern: $2 = 2 \times 1$, $6 = 3 \times 2$, $12 = 4 \times 3$, $20 = 5 \times 4$
- Pattern: $n(n+1)$
So $T_n = n(n+1) \cdot x^{n-1}$
$T_5 = 5 \times 6 \times x^4 = 30x^4$
Answer: $30x^4$
Strategy 2: Sum Using Standard Formula
Problem: Find sum of first 5 terms: $1 + 3x + 5x^2 + 7x^3 + 9x^4$
Solution:
This is AGP with:
- $a = 1$ (first term of AP)
- $d = 2$ (common difference of AP)
- $r = x$ (common ratio)
- $n = 5$
Using formula:
$$S_5 = \frac{1 - [1+4(2)]x^5}{1-x} + \frac{2x(1-x^4)}{(1-x)^2}$$ $$= \frac{1 - 9x^5}{1-x} + \frac{2x(1-x^4)}{(1-x)^2}$$For numerical answer, we’d need value of $x$.
Answer: $S_5 = \frac{1 - 9x^5}{1-x} + \frac{2x(1-x^4)}{(1-x)^2}$
Strategy 3: Infinite Sum Application
Problem: Find $1 + 2 \times \frac{1}{3} + 3 \times \frac{1}{9} + 4 \times \frac{1}{27} + \cdots$
Solution:
This is AGP with $a = 1, d = 1, r = \frac{1}{3}$
Check convergence: $|r| = \frac{1}{3} < 1$ ✓
$$S_\infty = \frac{1}{(1-\frac{1}{3})^2} = \frac{1}{(\frac{2}{3})^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}$$Answer: $\frac{9}{4}$
Strategy 4: Multiply and Subtract Method
Problem: Find sum: $S = 1 + 3x + 5x^2 + 7x^3 + \cdots$ (infinite, $|x| < 1$)
Solution:
$S = 1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots$
$xS = x + 3x^2 + 5x^3 + 7x^4 + \cdots$
Subtract:
$S - xS = 1 + 2x + 2x^2 + 2x^3 + 2x^4 + \cdots$
$S(1-x) = 1 + 2x(1 + x + x^2 + \cdots)$
$S(1-x) = 1 + 2x \cdot \frac{1}{1-x}$
$S(1-x) = 1 + \frac{2x}{1-x}$
$S(1-x) = \frac{1-x+2x}{1-x} = \frac{1+x}{1-x}$
$S = \frac{1+x}{(1-x)^2}$
Verify: Using formula with $a=1, d=2, r=x$:
$S_\infty = \frac{1 + x(2-1)}{(1-x)^2} = \frac{1+x}{(1-x)^2}$ ✓
Answer: $\frac{1+x}{(1-x)^2}$
Practice Problems
Level 1: JEE Main Basics
Problem 1: Find the 6th term of AGP: $3, 6r, 9r^2, 12r^3, \ldots$
Solution
Identify: $a = 3, d = 3, r = r$
$T_6 = [3 + 5(3)] \cdot r^5 = 18r^5$
Answer: $18r^5$
Problem 2: Find sum: $1 + 2 \times \frac{1}{2} + 3 \times \frac{1}{4} + 4 \times \frac{1}{8} + \cdots$ (infinite)
Solution
AGP with $a = 1, d = 1, r = \frac{1}{2}$
$|r| = \frac{1}{2} < 1$ ✓
$S_\infty = \frac{1}{(1-\frac{1}{2})^2} = \frac{1}{(\frac{1}{2})^2} = \frac{1}{\frac{1}{4}} = 4$
Answer: 4
Problem 3: Is $2, 6, 18, 54, \ldots$ an AGP?
Solution
Check ratios: $\frac{6}{2} = 3$, $\frac{18}{6} = 3$, $\frac{54}{18} = 3$
This is a GP, not AGP! (constant ratio, not increasing/decreasing coefficients)
Answer: No, it’s a GP
Level 2: JEE Main/Advanced
Problem 4: Find the sum of first 4 terms of AGP: $1, 4x, 7x^2, 10x^3, \ldots$
Solution
$a = 1, d = 3, r = x, n = 4$
Method 1: Direct calculation $S_4 = 1 + 4x + 7x^2 + 10x^3$
Method 2: Using formula
$$S_4 = \frac{1 - [1+3(3)]x^4}{1-x} + \frac{3x(1-x^3)}{(1-x)^2}$$ $$= \frac{1 - 10x^4}{1-x} + \frac{3x(1-x^3)}{(1-x)^2}$$For specific $x$, we can evaluate. Generally, answer is as above.
Answer: $1 + 4x + 7x^2 + 10x^3$ or formula form
Problem 5: If $1 + 3x + 5x^2 + 7x^3 + \cdots = \frac{9}{4}$ (infinite series), find $x$.
Solution
AGP: $a = 1, d = 2, r = x$
$S_\infty = \frac{1 + x(2-1)}{(1-x)^2} = \frac{1+x}{(1-x)^2} = \frac{9}{4}$
$4(1+x) = 9(1-x)^2$
$4 + 4x = 9(1 - 2x + x^2)$
$4 + 4x = 9 - 18x + 9x^2$
$9x^2 - 22x + 5 = 0$
Using quadratic formula:
$x = \frac{22 \pm \sqrt{484 - 180}}{18} = \frac{22 \pm \sqrt{304}}{18} = \frac{22 \pm 4\sqrt{19}}{18} = \frac{11 \pm 2\sqrt{19}}{9}$
Check convergence: We need $|x| < 1$
$x_1 = \frac{11 + 2\sqrt{19}}{9} \approx \frac{11 + 8.72}{9} \approx 2.19 > 1$ ✗
$x_2 = \frac{11 - 2\sqrt{19}}{9} \approx \frac{11 - 8.72}{9} \approx 0.25 < 1$ ✓
Answer: $x = \frac{11 - 2\sqrt{19}}{9}$
Problem 6: Find sum: $1 - 3x + 5x^2 - 7x^3 + \cdots$ where $|x| < 1$
Solution
This is AGP with alternating signs: $r = -x$
$a = 1, d = 2, r = -x$
$S_\infty = \frac{1 + (-x)(2-1)}{(1-(-x))^2} = \frac{1-x}{(1+x)^2}$
Answer: $\frac{1-x}{(1+x)^2}$
Level 3: JEE Advanced
Problem 7: Prove that $\sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2}$ for $|x| < 1$.
Solution
Method 1: Recognize as derivative
We know: $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ for $|x| < 1$
Differentiate both sides with respect to $x$:
$\sum_{k=1}^{\infty} kx^{k-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{0 \cdot (1-x) - 1 \cdot (-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$ ✓
Method 2: AGP approach
$S = 1 + 2x + 3x^2 + 4x^3 + \cdots$
This is AGP with $a = 1, d = 1, r = x$
$S_\infty = \frac{1}{(1-x)^2}$ ✓
Answer: Proven
Problem 8: Find $\sum_{k=1}^{n} k(k+1)x^k$ in closed form.
Solution
Expand: $k(k+1) = k^2 + k$
$\sum k(k+1)x^k = \sum k^2 x^k + \sum kx^k$
We know:
- $\sum_{k=1}^{\infty} kx^k = \frac{x}{(1-x)^2}$ (multiply standard AGP by $x$)
For $\sum k^2 x^k$, we can derive by differentiating $\sum kx^k$:
$\sum kx^k = \frac{x}{(1-x)^2}$
Differentiate: $\sum k^2x^{k-1} = \frac{d}{dx}\left(\frac{x}{(1-x)^2}\right)$
$= \frac{(1-x)^2 - x \cdot 2(1-x)(-1)}{(1-x)^4}$
$= \frac{(1-x)^2 + 2x(1-x)}{(1-x)^4}$
$= \frac{(1-x)(1-x+2x)}{(1-x)^4}$
$= \frac{1+x}{(1-x)^3}$
So: $\sum k^2 x^k = x \cdot \frac{1+x}{(1-x)^3} = \frac{x(1+x)}{(1-x)^3}$
Therefore:
$\sum k(k+1)x^k = \frac{x(1+x)}{(1-x)^3} + \frac{x}{(1-x)^2}$
$= \frac{x(1+x) + x(1-x)}{(1-x)^3}$
$= \frac{x(1+x+1-x)}{(1-x)^3}$
$= \frac{2x}{(1-x)^3}$
For finite sum to $n$ terms, this gets complex. The infinite case is cleaner.
Answer (infinite): $\frac{2x}{(1-x)^3}$ for $|x| < 1$
Important Results to Remember
1. Basic AGP Infinite Sum
$$\boxed{1 + 2x + 3x^2 + 4x^3 + \cdots = \frac{1}{(1-x)^2} \text{ for } |x| < 1}$$2. Shifted AGP
$$\boxed{x + 2x^2 + 3x^3 + 4x^4 + \cdots = \frac{x}{(1-x)^2} \text{ for } |x| < 1}$$3. Sum of Squares AGP
$$\boxed{1 + 4x + 9x^2 + 16x^3 + \cdots = \frac{1+x}{(1-x)^3} \text{ for } |x| < 1}$$4. General AGP
$$\boxed{a + (a+d)r + (a+2d)r^2 + \cdots = \frac{a + r(d-a)}{(1-r)^2} \text{ for } |r| < 1}$$Wait, let me recalculate this using the standard formula:
$S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$
$= \frac{a(1-r) + dr}{(1-r)^2}$
$= \frac{a - ar + dr}{(1-r)^2}$
$= \frac{a + r(d-a)}{(1-r)^2}$ ✓
Cross-Topic Connections
1. Link to Calculus - Derivatives
The formula $\sum kx^{k-1} = \frac{1}{(1-x)^2}$ is actually the derivative of $\sum x^k = \frac{1}{1-x}$!
This connects series summation to differentiation.
2. Link to Probability
Expected value calculations in geometric distributions often involve AGP.
3. Link to Infinite GP
AGP generalizes infinite GP by adding linear growth to exponential decay. See Infinite GP.
4. Link to Binomial Theorem
Differentiation of binomial series $(1+x)^n$ produces AGP-like patterns. See Binomial Theorem.
Quick Revision Checklist
- AGP = (AP term) × (GP term): $[a+(n-1)d] \cdot r^{n-1}$
- Infinite sum: $S_\infty = \frac{a + r(d-a)}{(1-r)^2}$ when $|r| < 1$
- Standard: $1 + 2x + 3x^2 + \cdots = \frac{1}{(1-x)^2}$
- Multiply and subtract method works well
- Always check $|r| < 1$ for infinite sums
- Watch for alternating signs (negative $r$)
- Connection to derivatives of power series
Memory Palace Technique
Imagine a factory assembly line (AGP):
- Workers (AP): Workers numbered $1, 2, 3, 4, \ldots$ (arithmetic)
- Productivity (GP): Each works at efficiency $1, r, r^2, r^3, \ldots$ (geometric)
- Output (AGP): Total output = worker number × efficiency: $1 \cdot 1, 2 \cdot r, 3 \cdot r^2, \ldots$
- Total Production: $\frac{1}{(1-r)^2}$ when efficiency ratio $r < 1$ (diminishing returns!)
- Formula Board: Shows $\frac{a + r(d-a)}{(1-r)^2}$ for general case
Final Tips for JEE
- Recognize pattern - increasing/decreasing coefficients with powers
- Check convergence - $|r| < 1$ mandatory for infinite sums
- Use multiply-subtract - often simpler than direct formula
- Remember $\frac{1}{(1-x)^2}$ - most common AGP result
- Connect to derivatives - $\sum kx^{k-1}$ is derivative of $\sum x^k$
- Watch alternating signs - indicates negative $r$
- Verify with small terms - calculate first few terms to check pattern
- Finite sum is complex - prefer infinite sum problems when possible
Last Updated: October 31, 2025