Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Chapter Introduction
Sequences and Series is Chapter 8 of CBSE Class 11 Mathematics, falling under the Algebra unit. This chapter extends the idea of number patterns you saw in earlier classes and introduces powerful tools for summing terms. Understanding AP and GP is essential because they appear directly in board exams (typically 6-8 marks) and form the foundation for series in competitive exams like JEE.
A sequence is an ordered arrangement of numbers generated by a definite rule. When you add the terms of a sequence, you get a series. This chapter covers arithmetic progressions, geometric progressions, arithmetic and geometric means, the AM-GM inequality, and summing special series.
Key Definitions
| Term | Definition |
|---|---|
| Sequence | An ordered list of numbers where each term follows a definite rule |
| Finite Sequence | A sequence with a limited number of terms (e.g., 1, 3, 5, 7, 9) |
| Infinite Sequence | A sequence that continues indefinitely (e.g., 1, 2, 3, 4, ...) |
| Series | The sum of terms of a sequence (e.g., 1 + 3 + 5 + 7 + 9) |
| Arithmetic Progression (AP) | A sequence where each term differs from the previous by a constant called common difference (d) |
| Geometric Progression (GP) | A sequence where each term is obtained by multiplying the previous term by a constant called common ratio (r) |
| Arithmetic Mean (AM) | The number inserted between two numbers such that the three form an AP |
| Geometric Mean (GM) | The positive number inserted between two positive numbers such that the three form a GP |
| AGP | A sequence where each term is the product of corresponding terms of an AP and a GP |
Detailed Concepts
1. Arithmetic Progression (AP)
An AP has the form: a, a + d, a + 2d, a + 3d, ...
nth term: a_n = a + (n - 1)d
Sum of first n terms: S_n = n/2 [2a + (n - 1)d] or S_n = n/2 (a + a_n)
Key properties of AP:
- If a constant is added to/subtracted from each term, the resulting sequence is also an AP.
- If each term is multiplied/divided by a non-zero constant, the resulting sequence is also an AP.
- The sum of terms equidistant from the beginning and end is constant: a_1 + a_n = a_2 + a_{n-1} = ...
- Three terms in AP can be taken as a - d, a, a + d
- Four terms in AP can be taken as a - 3d, a - d, a + d, a + 3d
Arithmetic Mean: Single AM between a and b is A = (a + b)/2. For n AMs between a and b, the common difference d = (b - a)/(n + 1).
2. Geometric Progression (GP)
A GP has the form: a, ar, ar^2, ar^3, ... (where r != 0)
nth term: a_n = ar^{n-1}
Sum of first n terms (r != 1): S_n = a(r^n - 1)/(r - 1) when r > 1; S_n = a(1 - r^n)/(1 - r) when r < 1
| **Sum of infinite GP ( | r | < 1):** S_infinity = a/(1 - r) |
|---|
Key properties of GP:
- If each term is multiplied/divided by a non-zero constant, the resulting sequence is also a GP.
- The reciprocals of a GP also form a GP.
- Three terms in GP can be taken as a/r, a, ar
- Four terms in GP can be taken as a/r^3, a/r, ar, ar^3
Geometric Mean: Single GM between a and b is G = sqrt(ab). For n GMs between a and b, the common ratio r = (b/a)^{1/(n+1)}.
3. AM-GM Inequality
For positive real numbers a_1, a_2, ..., a_n:
*(a_1 + a_2 + ... + a_n)/n >= (a_1 a_2 ... a_n)^{1/n}**
That is, AM >= GM, with equality if and only if a_1 = a_2 = ... = a_n.
For two numbers: (a + b)/2 >= sqrt(ab)
For three numbers: (a + b + c)/3 >= (abc)^{1/3}
4. Sum of Special Series
| Series | Formula |
|---|---|
| Sum of first n natural numbers | Sigma n = n(n + 1)/2 |
| Sum of squares of first n natural numbers | Sigma n^2 = n(n + 1)(2n + 1)/6 |
| Sum of cubes of first n natural numbers | Sigma n^3 = [n(n + 1)/2]^2 |
| Note: Sigma n^3 = (Sigma n)^2 | The cube sum equals the square of the natural number sum |
5. Method of Difference
When the nth term of a series is not directly identifiable, use the method of difference: 1. Write T_1, T_2, T_3, ... separately. 2. Find the differences T_2 - T_1, T_3 - T_2, etc. 3. If the first differences form an AP, then T_n is a quadratic in n. 4. Let T_n = An^2 + Bn + C, substitute n = 1, 2, 3 to get three equations, and solve for A, B, C. 5. Then S_n = A Sigma n^2 + B Sigma n + C * n.
6. Arithmetic-Geometric Progression (AGP)
An AGP has terms of the form: a, (a + d)r, (a + 2d)r^2, ...
Each term is the product of corresponding terms of an AP and a GP. To sum an AGP: 1. Write S_n = a + (a + d)r + (a + 2d)r^2 + ... + [a + (n-1)d]r^{n-1} 2. Multiply both sides by r: rS_n = ar + (a + d)r^2 + ... + [a + (n-1)d]r^n 3. Subtract: S_n(1 - r) = a + dr + dr^2 + ... + dr^{n-1} - [a + (n-1)d]r^n 4. Simplify using GP sum formula.
AP vs GP Comparison Table
| Property | AP | GP | ||
|---|---|---|---|---|
| Definition | a, a + d, a + 2d, ... | a, ar, ar^2, ... | ||
| Common element | Common difference (d) | Common ratio (r) | ||
| nth term | a + (n - 1)d | ar^{n-1} | ||
| Sum (n terms) | n/2 [2a + (n - 1)d] | a(r^n - 1)/(r - 1) | ||
| Mean between a, b | (a + b)/2 | sqrt(ab) | ||
| Infinite sum | Diverges (unless d = 0) | a/(1 - r) if | r | < 1 |
| Three terms form | a - d, a, a + d | a/r, a, ar | ||
| Increases by | Addition of d | Multiplication by r |
Sum Formulae Quick Reference
| What to find | Formula | ||
|---|---|---|---|
| nth term of AP | a_n = a + (n - 1)d | ||
| Sum of n terms of AP | S_n = n/2 [2a + (n - 1)d] | ||
| nth term of GP | a_n = ar^{n-1} | ||
| Sum of n terms of GP (r > 1) | S_n = a(r^n - 1)/(r - 1) | ||
| Sum of n terms of GP (r < 1) | S_n = a(1 - r^n)/(1 - r) | ||
| Sum of infinite GP | S = a/(1 - r), | r | < 1 |
| Single AM of a, b | (a + b)/2 | ||
| Single GM of a, b | sqrt(ab) | ||
| n AMs between a, b | d = (b - a)/(n + 1) | ||
| n GMs between a, b | r = (b/a)^{1/(n+1)} | ||
| Sigma n | n(n + 1)/2 | ||
| Sigma n^2 | n(n + 1)(2n + 1)/6 | ||
| Sigma n^3 | [n(n + 1)/2]^2 |
Important Numericals
Example 1: Finding nth term and sum of AP
Find the 10th term and sum of first 20 terms of the AP: 3, 7, 11, 15, ...
Here a = 3, d = 4.
a_10 = a + 9d = 3 + 9(4) = 39
S_20 = 20/2 [2(3) + 19(4)] = 10[6 + 76] = 10 x 82 = 820
Example 2: Inserting arithmetic means
Insert 4 arithmetic means between 3 and 23.
Total terms = 6 (3, four means, 23). These form an AP.
d = (23 - 3)/(4 + 1) = 20/5 = 4
The means are: 7, 11, 15, 19
Example 3: Sum of infinite GP
Find the sum: 1 + 2/3 + 4/9 + 8/27 + ...
| Here a = 1, r = 2/3. Since | r | < 1, the infinite sum exists. |
|---|
S = a/(1 - r) = 1/(1 - 2/3) = 1/(1/3) = 3
Example 4: AM-GM inequality application
For the numbers 4 and 9: AM = (4 + 9)/2 = 6.5, GM = sqrt(4 x 9) = sqrt(36) = 6.
Verify: AM >= GM: 6.5 >= 6 (holds).
Example 5: Sum of special series
Find Sigma k^2 from k = 1 to 20.
Sigma n^2 = n(n + 1)(2n + 1)/6 = 20(21)(41)/6 = 2870
Common Mistakes to Avoid
1. Confusing sequence and series: A sequence is a list; a series is the sum of that list. Do not mix them up in exam answers.
2. Wrong sign in sum formula: When r < 1, use S_n = a(1 - r^n)/(1 - r). When r > 1, use S_n = a(r^n - 1)/(r - 1). Both give the same result, but pick the one that avoids negative numerators.
| 3. Forgetting condition for infinite GP: The formula S = a/(1 - r) only works when | r | < 1. Always state this condition before applying the formula. |
|---|
4. Incorrect number of means: If you insert n AMs between a and b, total terms are n + 2 (including a and b). The common difference is (b - a)/(n + 1), not (b - a)/n.
5. AM-GM without positive numbers: The AM-GM inequality applies only to positive real numbers. Never apply it to negative or zero values.
6. GP with negative ratio: When r is negative, the signs alternate. Track signs carefully in each term.
7. Sigma notation confusion: Sigma n^3 = (Sigma n)^2 is true, but Sigma n^2 != (Sigma n)^2. This is a frequent exam trap.
Exam Tips
- For 1-mark questions, be ready with: definitions of AP, GP, AM, GM; the condition |r| < 1 for infinite GP; and the AM-GM statement.
- For 3-mark questions, practice: finding nth term or sum of AP/GP, inserting means, and verifying AM >= GM.
- For 5-mark questions, practice: method of difference problems, AGP sum, and proofs involving Sigma n, Sigma n^2, Sigma n^3.
- Always write the formula first, then substitute values. This ensures partial marks even if calculation goes wrong.
- In word problems, identify the sequence type first (AP or GP), then list what is given and what is asked.
- When the problem says "insert n means", remember to count total terms correctly.
NCERT Exercise Problem Patterns
| Exercise | Topic | Typical Question Type |
|---|---|---|
| 8.1 | Sequences, AP nth term | Find nth term, identify AP, find d |
| 8.2 | AP sum | Find S_n, find n when S_n is given |
| 8.3 | GP nth term and sum | Find a_n, S_n, S_infinity |
| 8.4 | AM, GM, AM-GM | Insert means, verify AM >= GM |
| 8.5 | Special series, method of difference | Sigma n, Sigma n^2, Sigma n^3, AGP |
Derivations to Remember
Derivation 1: Sum of AP formula
S_n = a + (a + d) + (a + 2d) + ... + [a + (n-1)d]
Write in reverse: S_n = a_n + (a_n - d) + (a_n - 2d) + ... + [a_n - (n-1)d]
Add the two: 2S_n = n(a + a_n)
S_n = n/2(a + a_n) = n/2[2a + (n-1)d]
Derivation 2: Sum of GP formula
S_n = a + ar + ar^2 + ... + ar^{n-1}
Multiply by r: rS_n = ar + ar^2 + ... + ar^n
Subtract: S_n(1 - r) = a - ar^n = a(1 - r^n)
S_n = a(1 - r^n)/(1 - r) for r != 1
| When | r | < 1 and n -> infinity: r^n -> 0, so S = a/(1 - r) |
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Derivation 3: AM-GM inequality for two numbers
Let a and b be positive. Consider (sqrt(a) - sqrt(b))^2 >= 0
Expanding: a - 2*sqrt(ab) + b >= 0
Equality when sqrt(a) = sqrt(b), i.e., a = b.
Additional Solved Examples
Example 6: Finding n when sum is given
How many terms of the AP 3, 8, 13, 18, ... must be taken so that their sum is 135?
a = 3, d = 5, S_n = 135
135 = n/2[2(3) + (n-1)(5)]
270 = n(6 + 5n - 5) = n(5n + 1)
5n^2 + n - 270 = 0
n = (-1 + sqrt(1 + 5400))/10 = (-1 + sqrt(5401))/10
Since n must be a positive integer, n = 9 (verify: 5(81) + 9 = 414, not 270). Let me recompute.
Actually: 270 = 5n^2 + n, so 5n^2 + n - 270 = 0. n = (-1 + sqrt(1 + 5400))/10 = (-1 + sqrt(5401))/10.
Approximately sqrt(5401) ~ 73.5, so n ~ 72.5/10 = 7.25. Check n = 9: S_9 = 9/2[6 + 40] = 9/2(46) = 207. Check n = 6: S_6 = 6/2[6 + 25] = 3(31) = 93. The answer requires checking integer values.
Example 7: GP sum to infinity
Find the sum to infinity: 1 - 1/2 + 1/4 - 1/8 + ...
| a = 1, r = -1/2. Since | r | = 1/2 < 1, the sum exists. |
|---|
S = a/(1 - r) = 1/(1 - (-1/2)) = 1/(3/2) = 2/3
Example 8: Using Sigma n^3 identity
Find the sum of cubes of the first 15 natural numbers.
Sigma n^3 = [n(n+1)/2]^2 = [15(16)/2]^2 = [120]^2 = 14400
Practice Checklist
- Solve at least 5 AP problems (nth term and sum)
- Solve at least 5 GP problems (nth term, sum, and infinite sum)
- Practice 2 AM/GM insertion problems each
- Solve 2 special series problems (Sigma n, n^2, n^3)
- Attempt 1 method of difference problem
- Verify AM >= GM for at least 3 pairs of numbers
Quick Revision Notes
1. AP: a_n = a + (n-1)d; S_n = n/2[2a + (n-1)d]; AM = (a+b)/2
3. AM >= GM always for positive numbers; equality when all numbers equal 4. Sigma n = n(n+1)/2; Sigma n^2 = n(n+1)(2n+1)/6; Sigma n^3 = [n(n+1)/2]^2 5. Method of difference: Find T_n, then sum using Sigma formulae 6. AGP: Multiply by r, subtract, use GP sum to simplify 7. n AMs between a and b: d = (b-a)/(n+1); n GMs between a and b: r = (b/a)^{1/(n+1)} 8. Three terms in AP: a-d, a, a+d. Three terms in GP: a/r, a, ar 9. In AP, sum of equidistant terms from start and end = a_1 + a_n (constant)
| 2. GP: a_n = ar^{n-1}; S_n = a(r^n-1)/(r-1); S_inf = a/(1-r) for | r | <1; GM = sqrt(ab) |
|---|---|---|
| 10. Always check | r | < 1 before using infinite GP sum formula |