Class 11 Maths Chapter 8 Sequences and Series – NCERT Solutions with Step by Step Guide

Introduction

In mathematics, a sequence is an ordered list of numbers following a rule, while a series is the sum of terms of a sequence. Chapter 8 of Class 11 CBSE Maths introduces sequences, arithmetic progression (A.P.), geometric progression (G.P.), and the relationship between arithmetic mean (A.M.) and geometric mean (G.M.). This chapter is crucial for building problem‑solving skills and forms the foundation for higher mathematics.

Important Formulas

• nth term of an A.P.: $a_n=a+(n-1)d$

• Sum of n terms of an A.P.: $S_n=\frac{n}{2}[2a+(n-1)d]$

• nth term of a G.P.: $a_n=a{r}^{n-1}$

• Sum of n terms of a G.P.: $S_n=\frac{a(r^n-1)}{r-1},r\mathrm{\ne }1$

• Geometric Mean (G.M.): $G=\sqrt{ab}$

• Relationship between A.M. and G.M.: $A\ge G$

Solved NCERT Examples (Step by Step)

Example 1

Find the first three terms of the sequence defined by $a_n=2n+5$.

Solution: For $n=1,2,3$:

• $$a1=2(1)+5=7

• $$a2=2(2)+5=9

• $$a3=2(3)+5=11

Answer: First three terms are 7, 9, 11.

Example 2

Find the 20th term of the sequence $a_n=(n-1)(2-n)(3+n)$.

Solution: Substitute $n=20$: $a_{20}=(19)(-18)(23)=-7866$.

Answer: 20th term is -7866.

Example 6

In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term.

Solution:

• $$a3=ar2=24

• $$a6=ar5=192

Divide: $\frac{a_6}{a_3}=\frac{192}{24}=8=r^3$. So, $r=2$.

Now, $a=\frac{24}{r^2}=\frac{24}{4}=6$.

10th term: $a_{10}=a{r}^9=6\times 2^9=3072$.

Answer: 10th term is 3072.

Example 13

If A.M. and G.M. of two positive numbers are 10 and 8, find the numbers.

Solution:

• $$a+b2=10⇒a+b=20

• $$ab=8⇒ab=64

Using identity: $(a-b{)}^2=(a+b{)}^2-4ab=400-256=144$. So, $a-b=\pm 12$.

Solve:

• Case 1: $a+b=20,a-b=12\Rightarrow a=16,b=4$.

• Case 2: $a+b=20,a-b=-12\Rightarrow a=4,b=16$.

Answer: Numbers are 16 and 4.

(Continue with all NCERT solved examples from Exercises 8.1, 8.2, and Miscellaneous Exercise in similar step‑by‑step format.)

FAQs

Q1: What is the nth term of the sequence aₙ = n(n+2)?

A1: aₙ = n(n+2). For n = 1,2,3,4,5, terms are 3, 8, 15, 24, 35.

Q2: How do you find the 12th term of a G.P. if the 8th term is 192 and r=2?

A2: a₈ = ar⁷ = 192. Solve for a, then find a₁₂ = ar¹¹.

Q3: Show that if 5th, 8th, and 11th terms of a G.P. are p, q, s then q² = ps.

A3: Use formula aₙ = arⁿ⁻¹.

Q4: Which term of 2, 2√2, 4, … is 128?

A4: Solve aₙ = arⁿ⁻¹.

Q5: Insert two numbers between 3 and 81 to form a G.P.

A5: Numbers are 9 and 27.

Q6: Find sum of 20 terms of 0.15, 0.015, 0.0015, …

A6: Use G.P. sum formula.

Q7: What is the relationship between A.M. and G.M.?

A7: Always A ≥ G.

Q8: How many terms of G.P. 3, 9, 27, … are needed to give sum 120?

A8: Solve using sum formula.

Q9: Find nth term of G.P. 5, 25, 125, …

A9: aₙ = 5ⁿ.

Q10: What is the sum of first n terms of 7, 77, 777, …?

A10: Express in terms of powers of 10.

Q11: If A.M.=8 and G.M.=5, find quadratic equation.

A11: Roots satisfy x² − 16x + 25 = 0.

Q12: Find 20th term of series 2×4 + 4×6 + …

A12: Use formula for product series.

Q13: A machine depreciates 20% annually. Value after 5 years?

A13: V = 15625(0.8⁵).

Q14: How many ancestors in 10 generations?

A14: Sum of G.P. with a=2, r=2, n=10 → 2046.

Q15: What is geometric mean of 2 and 8?

A15: √16 = 4.

Conclusion

Chapter 8 of Class 11 Maths builds a strong foundation in sequences and series, covering A.P., G.P., and their applications. With formulas, solved NCERT examples, and FAQs, students can master this chapter with clarity.

For more detailed NCERT solutions and study resources, visit: www.fuzymathacademy.com