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MathematicsClass 11Sequences and Series

Sequences and Series | Class 11 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Sequences and Series | Class 11 Mathematics Notes

Sequences and Series – this guide gives you a concise, exam-ready overview of Sequences and Series from Class 11 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

8.3 Series

This section introduces the concept of series as the sum of terms of a sequence. Given a sequence a_1, a_2, a_3, ..., a_n, the series is expressed as a_1 + a_2 + a_3 + ... + a_n + ..., which can be finite or infinite depending on the sequence. The compact sigma notation ∑_{k=1}^n a_k is introduced to represent the sum of the first n terms. The section clarifies the distinction between a series (the expression) and the sum of the series (the numerical value obtained by adding the terms). Several examples illustrate finding terms of sequences defined by formulas and writing corresponding series. For example, for a_n = 2n + 5, the first three terms are 7, 9, and 11. Another example involves a sequence defined by a recurrence relation, where a_1 = 1 and a_n = a_{n-1} + 2 for n ≥ 2, yielding terms 1, 3, 5, 7, 9 and the series 1 + 3 + 5 + 7 + 9 + ... . The section emphasizes understanding the general term and the sum of terms as foundational for further study of arithmetic and geometric progressions.

📊 Diagram: No diagrams in this section.

🧪 Activity: Exercises 8.1 involve writing first five terms of sequences defined by various formulas and finding indicated terms.

🔗 Connection: Understanding series prepares for the study of specific types of sequences and their sums, such as geometric progressions in the next section.

Frequently asked questions

Exercises 1 to 6 whose nth terms are: 1. an = n(n + 2) 2. an = n/(n + 1) 3. an = 2^n 4. an = (2n - 3)/6 5. an = (-1)^{n-1} * 5^{n+1} 6. an = n * (n^2 + 5)/4 Find the indicated terms in each of the sequences in Exercises 1 to 6.

For each sequence given by the nth term an, substitute the required values of n to find the indicated terms.

Example for sequence 1: an = n(n + 2)

  • To find a1, substitute n=1: a1 = 1*(1+2) = 3
  • To find a2, substitute n=2: a2 = 2*(2+2) = 8

Similarly, for each sequence, substitute the given n values to find the terms.

Write the first five terms of each of the sequences whose nth terms are: (i) a_n = 4n - 3 (ii) a_n = \frac{n^2}{2^n} (iii) a_n = (-1)^{n-1} n^3 (iv) a_n = \frac{n(n - 2)}{n + 3}

Solution: (i) a_n = 4n - 3 First five terms: For n=1: a_1 = 4(1) - 3 = 4 - 3 = 1 For n=2: a_2 = 4(2) - 3 = 8 - 3 = 5 For n=3: a_3 = 4(3) - 3 = 12 - 3 = 9 For n=4: a_4 = 4(4) - 3 = 16 - 3 = 13 For n=5: a_5 = 4(5) - 3 = 20 - 3 = 17 So, first five terms are: 1, 5, 9, 13, 17

(ii) a_n = \frac{n^2}{2^n} First five terms: For n=1: a_1 = 1^2 / 2^1 = 1/2 = 0.5 For n=2: a_2 = 4 / 4 = 1 For n=3: a_3 = 9 / 8 = 1.125 For n=4: a_4 = 16 / 16 = 1 For n=5: a_5 = 25 / 32 = 0.78125 So, first five terms are: 0.5,

11. $a_1 = 3, a_n = 3a_{n-1} + 2$ for all $n > 1$

Given the recurrence relation $a_1 = 3$, and $a_n = 3a_{n-1} + 2$ for $n > 1$, find the corresponding series.

Solution: Start with $a_1 = 3$. Calculate next terms: $a_2 = 3a_1 + 2 = 3 imes 3 + 2 = 11$ $a_3 = 3a_2 + 2 = 3 imes 11 + 2 = 35$ $a_4 = 3a_3 + 2 = 3 imes 35 + 2 = 107$ $a_5 = 3a_4 + 2 = 3 imes 107 + 2 = 323$

Thus, the series is: 3, 11, 35, 107, 323, ...

12. $a_1 = -1, a_n = \frac{a_{n-1}}{n}, n \geq 2$

Given $a_1 = -1$ and $a_n = \frac{a_{n-1}}{n}$ for $n \geq 2$, find the corresponding series.

Solution: Start with $a_1 = -1$. Calculate next terms: $a_2 = \frac{a_1}{2} = \frac{-1}{2} = -\frac{1}{2}$ $a_3 = \frac{a_2}{3} = \frac{-\frac{1}{2}}{3} = -\frac{1}{6}$ $a_4 = \frac{a_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24}$ $a_5 = \frac{a_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120}$

Thus, the series is: $-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}, ...$

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