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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.

8.2 Sequences

This section elaborates on the concept of sequences by providing examples and formal definitions. It begins with a practical example of calculating the number of ancestors over 300 years assuming a generation gap of 30 years, resulting in 10 generations. The number of ancestors in each generation forms a sequence: 2, 4, 8, 16, ..., 1024. Another example is the sequence of successive quotients obtained by dividing 10 by 3 repeatedly, yielding 3, 3.3, 3.33, 3.333, and so on. The section defines the terms of a sequence as a_1, a_2, ..., a_n, where the subscript denotes the position or order of the term. The nth term, a_n, is also called the general term. Sequences can be finite or infinite depending on the number of terms. The section explains that sequences often have algebraic formulas for their terms, such as the even natural numbers sequence where a_n = 2n, or the odd natural numbers sequence where a_n = 2n - 1. It also introduces sequences defined by recurrence relations, exemplified by the Fibonacci sequence where each term is the sum of the two preceding terms. The section notes that some sequences, like prime numbers, do not have explicit formulas but can be described verbally. Finally, it emphasizes that sequences can be viewed as functions with natural numbers as their domain, sometimes denoted as a(n).

📊 Diagram: No diagrams in this section.

🔗 Connection: This section's foundation on sequences leads to the concept of series, which is the sum of sequence terms, discussed 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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