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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.

8.4 Geometric Progression (G. P.)

This section defines geometric progression (G.P.) as a sequence where each term after the first is obtained by multiplying the preceding term by a constant non-zero number called the common ratio (r). Examples include sequences like 2, 4, 8, 16,... (common ratio 2), 1/9, -1/27, 1/81, -1/243,... (common ratio -1/3), and 0.01, 0.0001, 0.000001,... (common ratio 0.01). The section formalizes the definition: a sequence a_1, a_2, a_3,... is a G.P. if a_{k+1}/a_k = r for all k ≥ 1. The first term is denoted by a. The general term of a G.P. is derived as a_n = a × r^{n-1}. The section also introduces the sum of the first n terms of a G.P., denoted S_n. Two cases are considered: when r = 1, the sum is simply na; when r ≠ 1, the sum is S_n = a(r^n - 1)/(r - 1). Several examples illustrate finding terms and sums of G.P.s, including solving for unknown terms or number of terms given sums. The section also discusses the geometric mean (G.M.) of two positive numbers a and b, defined as √(ab), and shows how to insert geometric means between two numbers to form a G.P. The relationship between arithmetic mean (A.M.) and geometric mean (G.M.) is introduced, proving that A.M. ≥ G.M. with equality only when the two numbers are equal.

📊 Diagram: No diagrams in this section.

🧪 Activity: Exercises 8.2 provide problems on finding terms, sums, and properties of G.P.s, including proofs and applications.

🔗 Connection: This section's concepts of G.P. and sums lead to exploring the relationship between arithmetic mean and geometric mean 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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