Sequence And Series: CBSE Class 11 Maths NCERT Guide

Fundamental Concepts: Sequence, Series, and Progression

Sequence

A sequence is a collection of numbers arranged in a definite order according to some rule. The numbers in the sequence are called its 'terms'. For example, 2, 4, 6, 8,... is a sequence where each term is 2 more than the previous one.

Series

A series is the sum of the terms of a sequence. If a₁, a₂, a₃, ... is a sequence, then the corresponding series is a₁ + a₂ + a₃ + ... . For example, the series for the sequence 2, 4, 6, 8 is 2 + 4 + 6 + 8.

Progression

A progression is a special type of sequence where the terms follow a specific mathematical pattern. Arithmetic Progression (AP) and Geometric Progression (GP) are the two main types of progressions you will study.

Finite vs. Infinite

A sequence or series is 'finite' if it has a limited number of terms. It is 'infinite' if it continues forever.

Deep Dive: Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence where the difference between any two consecutive terms is constant. This constant difference is called the 'common difference' and is denoted by 'd'.

Example: The sequence 5, 8, 11, 14, 17, ... is an AP because the difference between consecutive terms is always 3. Here, the first term (a) is 5 and the common difference (d) is 3.

The n-th Term of an AP:
How do we find any term without listing them all out? Let the first term be 'a' and the common difference be 'd'.

• 1st term (a₁): a
• 2nd term (a₂): a + d
• 3rd term (a₃): a + 2d
• 4th term (a₄): a + 3d

Notice a pattern? The n-th term (aₙ) has (n-1) times the common difference added to the first term. This gives us the general formula:
aₙ = a + (n-1)d

Sum of the First n Terms of an AP (Sₙ):
To find the sum of the first n terms, we can use a clever trick. Let's write the sum forwards and backwards:
Sₙ = a + (a+d) + ... + (a + (n-1)d)
Sₙ = (a + (n-1)d) + (a + (n-2)d) + ... + a
Adding these two equations, each pair of terms sums to 2a + (n-1)d. Since there are 'n' such pairs:
2Sₙ = n * [2a + (n-1)d]
This gives us the formula for the sum:
Sₙ = n/2 [2a + (n-1)d]
An alternative version uses the last term (l = aₙ):
Sₙ = n/2 [a + l]

Understanding Geometric Progression (GP)

A Geometric Progression (GP) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the 'common ratio', denoted by 'r'.

Example: The sequence 2, 6, 18, 54, ... is a GP. Here, the first term (a) is 2 and the common ratio (r) is 3 (since 6/2 = 3, 18/6 = 3, etc.).

The n-th Term of a GP:
Let the first term be 'a' and the common ratio be 'r'.

• 1st term (a₁): a
• 2nd term (a₂): ar
• 3rd term (a₃): ar²
• 4th term (a₄): ar³

The pattern shows that the n-th term involves 'r' raised to the power of (n-1). The general formula is:
aₙ = arⁿ⁻¹

Sum of the First n Terms of a GP (Sₙ):
The sum Sₙ is a + ar + ar² + ... + arⁿ⁻¹. Multiplying by 'r', we get rSₙ = ar + ar² + ... + arⁿ. Subtracting the two:
Sₙ - rSₙ = a - arⁿ
Sₙ(1-r) = a(1-rⁿ)
This gives the sum formula (for r ≠ 1):
Sₙ = a(1 - rⁿ) / (1 - r) or a(rⁿ - 1) / (r - 1)

Sum of an Infinite GP:
What if the series goes on forever? If the common ratio 'r' is between -1 and 1 (i.e., |r| < 1), the term rⁿ gets closer and closer to 0 as n becomes very large. In this case, the sum converges to a finite value:
S∞ = a / (1 - r)

Worked Examples: AP and GP in Action

• Example 1 (AP): For the AP: 3, 7, 11, 15, ..., find the 18th term and the sum of the first 25 terms. Solution: Step 1: Identify the parameters. First term, a = 3. Common difference, d = 7 - 3 = 4. Step 2: Find the 18th term (a₁₈). Using the formula aₙ = a + (n-1)d: a₁₈ = 3 + (18 - 1) 4 a₁₈ = 3 + 17 4 = 3 + 68 = 71. Step 3: Find the sum of the first 25 terms (S₂₅). Using the formula Sₙ = n/2 [2a + (n-1)d]: S₂₅ = 25/2 [23 + (25 - 1)4] S₂₅ = 25/2 [6 + 244] = 25/2 [6 + 96] = 25/2 [102] S₂₅ = 25 * 51 = 1275. Final Answer: The 18th term is 71 and the sum of the first 25 terms is 1275.
• Example 2 (GP): Find the sum of the first 8 terms of the GP: 5, 10, 20, 40, ... Solution: Step 1: Identify the parameters. First term, a = 5. Common ratio, r = 10/5 = 2. Number of terms, n = 8. Step 2: Apply the sum formula for a GP. Since r > 1, we use Sₙ = a(rⁿ - 1) / (r - 1). S₈ = 5 (2⁸ - 1) / (2 - 1) Step 3: Calculate the result. S₈ = 5 (256 - 1) / 1 S₈ = 5 255 = 1275. Final Answer: The sum of the first 8 terms is 1275.
• Example 3 (Infinite GP): Find the sum of the infinite GP: 16, 8, 4, 2, ... Solution: Step 1: Identify parameters and check for convergence. First term, a = 16. Common ratio, r = 8/16 = 1/2. Since |r| = 1/2 < 1, the sum to infinity exists. Step 2: Apply the infinite sum formula. Using S∞ = a / (1 - r): S∞ = 16 / (1 - 1/2) S∞ = 16 / (1/2) = 16 * 2 = 32. Final Answer: The sum of the infinite GP is 32.

Sum of Special Series

Besides AP and GP, there are some standard series whose sum formulas are very useful, especially when dealing with more complex problems. These are the sums of powers of the first 'n' natural numbers.

1. Sum of the first n natural numbers:

Σn = 1 + 2 + 3 + ... + n = n(n+1) / 2
(This is just an AP with a=1 and d=1)

1. Sum of the squares of the first n natural numbers:

Σn² = 1² + 2² + 3² + ... + n² = n(n+1)(2n+1) / 6

1. Sum of the cubes of the first n natural numbers:

Σn³ = 1³ + 2³ + 3³ + ... + n³ = [n(n+1) / 2]² = (Σn)²

These formulas are essential tools for solving problems that might not be a simple AP or GP but can be broken down into these components. For example, to find the sum of the series (12) + (23) + (3*4) + ... up to n terms, you can write the k-th term as k(k+1) = k² + k and then sum it from k=1 to n using the formulas above: Σ(k² + k) = Σk² + Σk.

Exam Traps & Key Pointers

Many students lose marks not because they don't know the concepts, but due to small mistakes. Here’s what to watch out for:

• Confusing AP and GP: Always check if there's a common difference (AP) or a common ratio (GP) before you start. Applying the wrong set of formulas is a very common error.
• Negative Common Ratio (r): Be extra careful with signs when 'r' is negative. For example, in the GP 3, -6, 12, ..., r = -2. When calculating rⁿ, remember that (-2)³ = -8 but (-2)⁴ = 16.
• Infinite GP Sum Condition: The formula S∞ = a / (1 - r) is ONLY valid if -1 < r < 1. If |r| ≥ 1, the sum diverges (goes to infinity) and cannot be calculated with this formula. Always check this condition first.
• n vs. n-1: A frequent slip-up is mixing up 'n' and 'n-1'. The n-th term has (n-1)d or rⁿ⁻¹. The sum formulas involve 'n' and rⁿ. Double-check which one you're using.
• Systematic Approach: Before solving, always write down the given values clearly: a = ?, d = ? or r = ?, n = ?. This reduces the chances of careless errors.

Practice Questions with Solutions

• Q: The 5th term of an AP is 19 and the 9th term is 35. Find the 15th term of the AP. A: Step 1: Set up equations using the formula aₙ = a + (n-1)d. a₅ = a + (5-1)d = a + 4d = 19 --- (i) a₉ = a + (9-1)d = a + 8d = 35 --- (ii) Step 2: Solve the simultaneous equations for 'a' and 'd'. Subtracting (i) from (ii): (a + 8d) - (a + 4d) = 35 - 19 4d = 16 => d = 4. Substitute d=4 into (i): a + 4(4) = 19 => a + 16 = 19 => a = 3. Step 3: Find the 15th term (a₁₅). Using a = 3 and d = 4: a₁₅ = a + (15-1)d = 3 + 14(4) = 3 + 56 = 59. Final answer: The 15th term is 59.
• Q: How many terms of the GP 3, 3/2, 3/4, ... are needed to give a sum of 3069/512? A: Step 1: Identify the parameters of the GP. First term, a = 3. Common ratio, r = (3/2) / 3 = 1/2. The given sum, Sₙ = 3069/512. Step 2: Use the sum formula for a GP, Sₙ = a(1 - rⁿ) / (1 - r). 3069/512 = 3(1 - (1/2)ⁿ) / (1 - 1/2) 3069/512 = 3(1 - (1/2)ⁿ) / (1/2) 3069/512 = 6(1 - (1/2)ⁿ) Step 3: Solve for 'n'. (3069/512) / 6 = 1 - (1/2)ⁿ 3069 / (512 * 6) = 1 - (1/2)ⁿ 1023 / 1024 = 1 - (1/2)ⁿ (1/2)ⁿ = 1 - 1023/1024 (1/2)ⁿ = 1/1024 Since 2¹⁰ = 1024, we have (1/2)ⁿ = (1/2)¹⁰. Therefore, n = 10. Final answer: 10 terms are needed.
• Q: A man saves ₹200 in the first month, ₹250 in the second, ₹300 in the third, and so on. How much will he save in 2 years? A: Step 1: Identify the type of progression. The savings form an AP: 200, 250, 300, ... First term, a = 200. Common difference, d = 50. Number of terms, n = 2 years = 24 months. Step 2: Apply the sum formula for an AP, Sₙ = n/2 [2a + (n-1)d]. We need to find S₂₄. S₂₄ = 24/2 [2(200) + (24-1)50] Step 3: Calculate the total savings. S₂₄ = 12 [400 + 23 * 50] S₂₄ = 12 [400 + 1150] S₂₄ = 12 [1550] = 18600. Final answer: He will save ₹18,600 in 2 years.
• Q: Find the sum of the series 1² + 2² + 3² + ... + 20². A: Step 1: Identify the series. This is the sum of the squares of the first 20 natural numbers. Step 2: Apply the formula for the sum of squares, Σn² = n(n+1)(2n+1) / 6. Here, n = 20. Sum = 20(20+1)(220+1) / 6 Sum = 20 21 41 / 6 Step 3: Calculate the result. Sum = (10 2) (3 7) 41 / (2 3) Sum = 10 7 41 Sum = 70 * 41 = 2870. Final answer: The sum of the series is 2870.

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