Sequences and Series: Class 11 Maths NCERT Guide

Fundamental Concepts: Sequence vs. Series

Sequence

A sequence is an ordered list of numbers, called terms, that follow a specific rule or pattern. Each term is denoted by a_n, where 'n' is its position in the sequence. For example, 3, 6, 9, 12,... is a sequence where each term is a multiple of 3.

Series

A series is the sum of the terms of a sequence. If a_1, a_2, a_3, ... is a sequence, then the corresponding series is a_1 + a_2 + a_3 + ... The sum of the first 'n' terms is denoted by S_n.

Progression

A progression is a special type of sequence where the terms follow a very specific mathematical formula, allowing us to predict any term in the sequence. Arithmetic Progressions (AP) and Geometric Progressions (GP) are key examples.

Arithmetic Progression (A.P.) in Depth

An Arithmetic Progression (A.P.) is a sequence where the difference between any two consecutive terms is constant. This constant value is called the common difference (d).

• General Form: An A.P. is written as: a, a+d, a+2d, a+3d, ... where a is the first term.

• The nth Term (a_n): To find any term in the sequence without writing it all out, we use the formula:

a_n = a + (n-1)d
Here, a_n is the term you want to find, a is the first term, n is the position of the term, and d is the common difference.

• Sum of the First n Terms (S_n): To find the sum of a part of the sequence, we have two useful formulas:

1. S_n = n/2 * [2a + (n-1)d] - Use this when you know the first term, common difference, and number of terms.
2. S_n = n/2 * [a + l] - Use this as a shortcut when you know the first term (a), the last term (l), and the number of terms (n). The intuition for this second formula is simple: if you pair the first and last terms, the second and second-to-last, and so on, each pair sums to a+l. There are n/2 such pairs.

Geometric Progression (G.P.) Explained

A Geometric Progression (G.P.) 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 (r).

• General Form: A G.P. is written as: a, ar, ar^2, ar^3, ... where a is the first term.

• The nth Term (a_n): The formula to find any specific term is:

a_n = a * r^(n-1)
Here, a is the first term, r is the common ratio, and n is the term's position.

• Sum of the First n Terms (S_n): The sum formula depends on the value of r:

S_n = a(r^n - 1) / (r - 1) when |r| > 1
S_n = a(1 - r^n) / (1 - r) when |r| < 1
Using the correct version helps avoid negative numbers in the denominator.

• Sum of an Infinite G.P. (S_∞): For a G.P. where the common ratio r is between -1 and 1 (i.e., |r| < 1), the sum of an infinite number of terms converges to a finite value. The formula is:

S_∞ = a / (1 - r)
This works because as n becomes infinitely large, r^n approaches zero, making the (1-r^n) part of the standard sum formula simply (1-0) = 1.

Worked Examples: Applying AP and GP Formulas

• Example 1 (A.P.): Find the 15th term and the sum of the first 20 terms of the A.P.: 3, 7, 11, 15, ... Solution: Step 1: Identify the first term (a) and common difference (d). The first term a = 3. The common difference d = 7 - 3 = 4. Step 2: Calculate the 15th term (a_15). Using the formula a_n = a + (n-1)d: a_15 = 3 + (15-1) 4 = 3 + 14 4 = 3 + 56 = 59. Step 3: Calculate the sum of the first 20 terms (S_20). Using the formula S_n = n/2 [2a + (n-1)d]: S_20 = 20/2 [23 + (20-1)4] = 10 [6 + 194] = 10 [6 + 76] = 10 * 82 = 820. Final Answer: The 15th term is 59 and the sum of the first 20 terms is 820.
• Example 2 (G.P.): In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term. Solution: Step 1: Set up equations using the nth term formula for a G.P., a_n = ar^(n-1). 3rd term: a_3 = ar^(3-1) = ar^2 = 24 ---(i) 6th term: a_6 = ar^(6-1) = ar^5 = 192 ---(ii) Step 2: Divide equation (ii) by equation (i) to find the common ratio (r). (ar^5) / (ar^2) = 192 / 24 r^3 = 8 r = 2. Step 3: Substitute r = 2 into equation (i) to find the first term (a). a (2)^2 = 24 a 4 = 24 a = 6. Step 4: Calculate the 10th term (a_10) using a=6 and r=2. a_10 = ar^(10-1) = ar^9 = 6 (2)^9 = 6 512 = 3072. Final Answer: The 10th term is 3072.

Important Formulas and Exam Traps

To excel in this chapter, memorize these formulas and be aware of common pitfalls.

Key Formula Summary:

• A.P. nth Term: a_n = a + (n-1)d
• A.P. Sum: S_n = n/2 [2a + (n-1)d] OR S_n = n/2 [a + l]
• G.P. nth Term: a_n = a * r^(n-1)
• G.P. Sum: S_n = a(r^n - 1) / (r - 1) for |r| > 1
• G.P. Infinite Sum: S_∞ = a / (1 - r) for |r| < 1

Common Exam Traps:

1. Confusing AP and GP: The first step in any problem should be to determine if there's a common difference (AP) or a common ratio (GP). Don't apply the wrong set of formulas.
2. Incorrect G.P. Sum Formula: Using S_n = a(r^n-1)/(r-1) when |r|<1 can lead to confusing negative signs. Stick to S_n = a(1-r^n)/(1-r) for |r|<1 to keep calculations straightforward.
3. Misusing the Infinite Sum Formula: The formula S_∞ = a/(1-r) is ONLY valid for a Geometric Progression and ONLY when -1 < r < 1. Applying it to an AP or a diverging GP (where |r| ≥ 1) is a major conceptual error.

Practice Questions with Solutions

• Q: How many terms of the A.P.: 24, 21, 18, ... must be taken so that their sum is 78? A: Step 1: Identify the properties of the A.P. The first term a = 24. The common difference d = 21 - 24 = -3. The sum S_n = 78. Step 2: Use the sum formula S_n = n/2 [2a + (n-1)d] and substitute the known values. 78 = n/2 [2(24) + (n-1)(-3)] 156 = n [48 - 3n + 3] 156 = n [51 - 3n] 156 = 51n - 3n^2 Step 3: Rearrange into a quadratic equation and solve for n. 3n^2 - 51n + 156 = 0 Divide by 3: n^2 - 17n + 52 = 0 Factoring the quadratic: (n-4)(n-13) = 0 So, n = 4 or n = 13. Final answer: Both answers are valid. The sum of the first 4 terms is 78, and the sum of the first 13 terms is also 78 (because the terms from 5th to 13th include positive and negative values that cancel each other out and also sum to zero with some other terms).
• Q: Find the sum of the first 8 terms of the G.P. 1, 3, 9, ... A: Step 1: Identify the properties of the G.P. The first term a = 1. The common ratio r = 3/1 = 3. The number of terms n = 8. Step 2: Since r = 3 which is greater than 1, use the G.P. sum formula S_n = a(r^n - 1) / (r - 1). Step 3: Substitute the values and calculate. S_8 = 1 * (3^8 - 1) / (3 - 1) S_8 = (6561 - 1) / 2 S_8 = 6560 / 2 = 3280. Final answer: The sum of the first 8 terms is 3280.
• Q: The sum of three numbers in an A.P. is 27 and their product is 504. Find the numbers. A: Step 1: Let the three numbers in A.P. be (a-d), a, (a+d). This is a standard trick that simplifies calculations. Step 2: Use the given sum to find the middle term 'a'. Sum: (a-d) + a + (a+d) = 27 3a = 27 a = 9. Step 3: Use the given product to find the common difference 'd'. We now know the numbers are (9-d), 9, (9+d). Product: (9-d) 9 (9+d) = 504 9 * (81 - d^2) = 504 81 - d^2 = 504 / 9 81 - d^2 = 56 d^2 = 81 - 56 = 25 d = ±5. Step 4: Find the three numbers for both values of d. If d=5, the numbers are (9-5), 9, (9+5), which are 4, 9, 14. If d=-5, the numbers are (9-(-5)), 9, (9-5), which are 14, 9, 4. Final answer: The numbers are 4, 9, and 14.
• Q: Find the sum to infinity of the geometric series: 5 + 5/2 + 5/4 + 5/8 + ... A: Step 1: Identify the first term a and the common ratio r. a = 5. r = (5/2) / 5 = 1/2. Step 2: Check if the sum to infinity formula can be applied. The condition is |r| < 1. Here, |1/2| < 1, so the formula is valid. Step 3: Apply the sum to infinity formula S_∞ = a / (1 - r). S_∞ = 5 / (1 - 1/2) S_∞ = 5 / (1/2) S_∞ = 5 * 2 = 10. Final answer: The sum to infinity of the series is 10.

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