Mastering AGP and Method of Differences for JEE Advanced
Complete guide to Arithmetic-Geometric Progression and Method of Differences with formulas, techniques, and problem-solving strategies.
Why AGP and Method of Differences Matter
These two methods are essential for solving complex series problems in JEE Advanced. Mastery of these techniques can help you secure 4-8 marks in every paper:
- AGP (Arithmetic-Geometric Progression) - Combines AP and GP for complex series
- Method of Differences - Powerful technique for telescoping series
- Time-saving approaches for complex-looking series problems
- Systematic problem-solving for any series variation
Arithmetic-Geometric Progression (AGP)
A sequence where each term is the product of corresponding terms of an AP and a GP.
General Form:
$S = a + (a+d)r + (a+2d)r^2 + \cdots + [a+(n-1)d]r^{n-1}$
Sum Formula:
For $r \neq 1$:
$$S_n = \frac{a - [a+(n-1)d]r^n}{1-r} + \frac{dr(1-r^{n-1})}{(1-r)^2}$$
Example: Find sum of $1 + 3x + 5x^2 + 7x^3 + \cdots$ to $n$ terms
Step 1: Identify AP: $1, 3, 5, 7, \cdots$ (a=1, d=2)
Step 2: Identify GP: $1, x, x^2, x^3, \cdots$ (r=x)
Step 3: Apply AGP formula:
$$S_n = \frac{1 - [1+(n-1)2]x^n}{1-x} + \frac{2x(1-x^{n-1})}{(1-x)^2}$$
Step 4: Simplify:
$$S_n = \frac{1 - (2n-1)x^n}{1-x} + \frac{2x(1-x^{n-1})}{(1-x)^2}$$
Example: Sum of series $1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \cdots$ to infinity
Step 1: AP: $1, 2, 3, 4, \cdots$ (a=1, d=1)
Step 2: GP: $1, \frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \cdots$ (r=1/3)
Step 3: For infinite series ($|r| < 1$):
$$S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$$
Step 4: Substitute values:
$$S_\infty = \frac{1}{1-\frac{1}{3}} + \frac{1 \cdot \frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{3}{2} + \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$$
Method of Differences
A powerful technique where we express terms as differences of consecutive terms of another sequence.
When to Use:
Series where terms can be written as $T_k = f(k) - f(k+1)$ or $T_k = f(k+1) - f(k)$
Example: Find sum of $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}$
Step 1: Write general term: $T_k = \frac{1}{k(k+1)}$
Step 2: Express as difference: $T_k = \frac{1}{k} - \frac{1}{k+1}$
Step 3: Write the sum:
$$S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$
Step 4: Cancel intermediate terms:
$$S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$
Example: Sum of series $1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n(n+1)$
Step 1: General term: $T_k = k(k+1) = k^2 + k$
Step 2: Use known formulas:
$$\sum k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum k = \frac{n(n+1)}{2}$$
Step 3: Add both:
$$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1+3)}{6} = \frac{n(n+1)(n+2)}{3}$$
🚀 Advanced Problem-Solving Strategies
For AGP Problems:
- Always multiply by common ratio and subtract
- Watch for infinite series when |r| < 1
- Simplify final expression completely
- Check special cases (r=1, etc.)
For Method of Differences:
- Look for telescoping pattern
- Try partial fractions for rational terms
- Use known summation formulas when possible
- Verify with small values of n
🎯 JEE Advanced Level Problems
Problem 1: Find sum of series $1^2 + 2^2x + 3^2x^2 + 4^2x^3 + \cdots$ to infinity where $|x| < 1$
Hint: Consider AGP with terms $n^2x^{n-1}$
Problem 2: Sum of series $\frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4} + \frac{1}{3\cdot4\cdot5} + \cdots$ to $n$ terms
Hint: Use method of differences with partial fractions
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find sum: $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \cdots$ to infinity
2. Sum the series: $\frac{1}{1\cdot3} + \frac{1}{3\cdot5} + \frac{1}{5\cdot7} + \cdots + \frac{1}{(2n-1)(2n+1)}$
3. Find sum: $1\cdot2\cdot3 + 2\cdot3\cdot4 + 3\cdot4\cdot5 + \cdots + n(n+1)(n+2)$
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