CBSE Class 11 Maths Important Questions - Chapter 8 Sequences and Series

1 Mark Questions with Solutions

1. Define an arithmetic progression (AP).  

Solution: A sequence is called an AP if the difference between consecutive terms is constant. For example, in $2, 4, 6, 8, \dots$, the common difference is $2$.

2. Write the general term of a geometric progression (GP).  

Solution: The general term of a GP is given by $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

3. Find the 10th term of the AP: $3, 7, 11, \dots$.  

Solution: The general term of an AP is $a_n = a + (n-1)d$. Here, $a = 3$ and $d = 7 - 3 = 4$:

$a_{10} = 3 + (10-1) \cdot 4 = 3 + 36 = 39.$

4. If the first term of a GP is $5$ and the common ratio is $2$, write the first four terms.  

Solution: The terms of the GP are:

$5, \, 5 \cdot 2 = 10, \, 5 \cdot 2^2 = 20, \, 5 \cdot 2^3 = 40.$

5. Find the sum of the first $5$ natural numbers.  

Solution: The sum of the first $n$ natural numbers is $S_n = \dfrac{n(n+1)}{2}$. For $n = 5$:

$S_5 = \dfrac{5(5+1)}{2} = \dfrac{5 \cdot 6}{2} = 15.$

6. What is the common difference of the AP: $2, 5, 8, 11, \dots$?  

Solution: The common difference is:

$d = 5 - 2 = 3.$

7. Write the formula for the sum of the first $n$ terms of an AP.  

Solution: The sum of the first $n$ terms of an AP is given by:

$S_n = \dfrac{n}{2} [2a + (n-1)d].$

8. State whether the sequence $1, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8}, \dots$ is an AP or a GP.  

Solution: It is a GP because the ratio between consecutive terms is constant:

$r = \dfrac{\dfrac{1}{2}}{1} = \dfrac{1}{2}.$

9. Write the first term and the common ratio of the GP: $3, 6, 12, \dots$.  

Solution: The first term is $a = 3$, and the common ratio is $r = \dfrac{6}{3} = 2$.

10. Find the value of $a_{n+1} - a_n$ for the AP: $4, 9, 14, 19, \dots$.  

Solution: The difference between consecutive terms in an AP is the common difference:

$a_{n+1} - a_n = d = 9 - 4 = 5.$

2 Mark Questions with Solutions

1 Find the 20th term of the AP: $5, 10, 15, 20, \dots$.  

Solution: The general term of an AP is $a_n = a + (n-1)d$. Here, $a = 5$ and $d = 10 - 5 = 5$:

$a_{20} = 5 + (20-1) \cdot 5 = 5 + 95 = 100.$

2. $ The 5th term of an AP is $12$, and the 10th term is $27$. Find the common difference.  

Solution: The general term is $a_n = a + (n-1)d$. For the 5th term:

$a + 4d = 12 \quad \text{(1)}.$

For the 10th term:

$a + 9d = 27 \quad \text{(2)}$

Subtracting (1) from (2):

$5d = 15 \implies d = 3.$

3. $ Find the sum of the first $15$ terms of the AP: $7, 10, 13, \dots$.  

Solution: The sum of the first $n$ terms of an AP is:

$S_n = \dfrac{n}{2} [2a + (n-1)d]$

Here, $a = 7$, $d = 10 - 7 = 3$, and $n = 15$:

$S_{15} = \dfrac{15}{2} [2 \cdot 7 + (15-1) \cdot 3] = \dfrac{15}{2} [14 + 42] = \dfrac{15}{2} \cdot 56 = 420$

4. $ If the first term of a GP is $3$ and the common ratio is $4$, find the 5th term.  

Solution: The general term of a GP is $a_n = ar^{n-1}$. Here, $a = 3$ and $r = 4$:

$a_5 = 3 \cdot 4^{5-1} = 3 \cdot 4^4 = 3 \cdot 256 = 768$

5. $ Show that the sequence $5, 8, 11, 14, \dots$ is an AP.  

Solution: Find the differences between consecutive terms:

$8 - 5 = 3, \, 11 - 8 = 3, \, 14 - 11 = 3$  

Since the difference is constant, it is an AP with a common difference $d = 3$.

3 Marks Important Questions

1. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\dfrac{n}{n+1}\] .

Ans:

The given equation is \[{{a}_{n}}=\dfrac{n}{n+1}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=\dfrac{1}{1+1}\]

\[\Rightarrow {{a}_{1}}=\dfrac{1}{2}\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=\dfrac{2}{2+1}\]

\[\Rightarrow {{a}_{2}}=\dfrac{2}{3}\]

\[{{a}_{3}}=\dfrac{3}{3+1}\]

\[\Rightarrow {{a}_{3}}=\dfrac{3}{4}\]

\[{{a}_{4}}=\dfrac{4}{4+1}\]

\[\Rightarrow {{a}_{4}}=\dfrac{4}{5}\]

\[{{a}_{5}}=\dfrac{5}{5+1}\]

\[\Rightarrow {{a}_{5}}=\dfrac{5}{6}\]

Therefore, the first five terms of \[{{a}_{n}}=\dfrac{n}{n+1}\] is \[\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5}\] and \[\dfrac{5}{6}\] . 

2. Show that the products of the corresponding terms of the sequences form \[a,ar,a{{r}^{2}},...a{{r}^{n-1}}\] and \[A,AR,A{{R}^{2}},A{{R}^{n-1}}\]  a G.P. and find the common ratio.

Ans:

The sequence \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\]  forms a G.P. is to be proved.

Second term / First term \[=\frac{arAR}{aA}=rR\]

Third term / Second term \[=\frac{a{{r}^{2}}A{{R}^{2}}}{aA}=rR\]

Therefore, the \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\] forms a G.P. and the common ratio is \[rR\].

3. What will Rs.\[500\] amounts to in \[10\] years after its deposit in a bank which pays annual interest rate of \[10%\] compounded annually?

Ans:

Rs.\[500\] is the amount deposited in the bank.

The amount \[=\] Rs.\[500\left( 1+\frac{1}{10} \right)=\] Rs.\[500\left( 1.1 \right)\] , at the end of first year.

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{2}^{nd}}\] year. 

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{3}^{rd}}\] year and so on. 

Therefore, the amount at the end of \[10\] years 

\[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)...\](\[10\]times)

\[=\] Rs.\[500{{\left( 1.1 \right)}^{10}}\]

4. The first term of a G.P. is \[1\] . The sum of the third term and fifth term is \[90\]. Find the common ratio of G.P.

Ans:

Let the first term of the G.P. be \[a\] and the common ratio be \[r\] .

Then, \[a=1\]

\[{{a}_{3}}=a{{r}^{2}}={{r}^{2}}\]

\[{{a}_{5}}=a{{r}^{4}}={{r}^{4}}\]

Therefore,

\[{{r}^{2}}+{{r}^{4}}=90\]

\[\Rightarrow {{r}^{4}}+{{r}^{2}}-90=0\]

\[\Rightarrow {{r}^{2}}=\frac{-1+\sqrt{1+360}}{2}\]

\[=\frac{-1+\sqrt{361}}{2}\]

\[=-10\] or \[9\]

\[\Rightarrow r=\pm 3\]

Therefore, \[\pm 3\] is the common ratio of the G.P. 

5. Find the \[{{20}^{th}}\] term of the series \[2\times 4+4\times 6+6\times 8+...+n\] terms.

Ans:

\[2\times 4+4\times 6+6\times 8+...+n\] is the given series,

Therefore the \[{{n}^{th}}\] term \[{{a}_{n}}=2n\times \left( 2n+2 \right)=4{{n}^{2}}+4n\]

Then,

\[{{a}_{20}}=4{{\left( 20 \right)}^{2}}+4\left( 20 \right)\]

\[=4\left( 400 \right)+80\]

\[=1600+80\]

\[=1680\]

Therefore, \[1680\] is the \[{{20}^{th}}\] term of the series. 

Long Answer Question (6 Mark)

1. 150 workers were engaged to finish a job in a certain no. of days. 4 workers dropped out on a second day, 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work and find the no. of days in which the work was completed?

Ans: A = 150, d = -4

${{S}_{n}}=~\frac{n}{2}\left[ 2\times 150+(n-1)(-4) \right]$

If total workers who would have worked for all n days, 150(n - 8)

$\therefore \dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{300}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)( - 4)}}} \right]\,{\text{ = }}\,{\text{150(n}}\,{\text{ - }}\,{\text{8)}}$

$ \Rightarrow \,{\text{n}}\,{\text{ = }}\,{\text{25}}$

2. Prove that the sum of n terms of the series ${\text{11}}\,{\text{ + }}\,{\text{103}}\,{\text{ + }}\,{\text{1005}}\,{\text{ + }}\,.....\,{\text{is}}\,\dfrac{{{\text{10}}}}{{\text{9}}}{\text{(1}}{{\text{0}}^{\text{n}}}\,{\text{ - }}\,{\text{1)}}\,{\text{ + }}\,{{\text{n}}^{\text{2}}}$.

Ans: ${\text{Sn  =  11  +  103  +  1005  +  }}......{\text{  +  n terms}}$

${\text{Sn  =  (10 + 1)  +  (1}}{{\text{0}}^{\text{2}}}{\text{ + 3)  +  (1}}{{\text{0}}^{\text{3}}}{\text{ + 5)  +  }}....{\text{  +  (10n}} + \,({\text{2n - 1}}){\text{)}}$

${\text{Sn  =  }}\dfrac{{{\text{10(1}}{{\text{0}}^{{\text{n}}\,}}{\text{ - }}\,{\text{1)}}}}{{{\text{10}}\,{\text{ - }}\,{\text{1}}}}\,{\text{ + }}\,\dfrac{{\text{n}}}{{\text{2}}}{\text{(1}}\,{\text{ + }}\,{\text{2n}}\,{\text{ - 1)}}$

$=\,\dfrac{{{\text{10}}}}{{\text{9}}}{\text{(1}}{{\text{0}}^{\text{n}}}\,{\text{ - }}\,{\text{1)}}\,{\text{ + }}\,{{\text{n}}^{\text{2}}}$

3. The ratio of A.M. and G.M. of two positive no. a and b is m : n. Show that

${\text{a : b  =  (m  +  }}\sqrt {{{\text{m}}^{\text{2}}}{\text{ - }}{{\text{n}}^{\text{2}}}} {\text{)}}\,{\text{:}}\,{\text{(m}}\,{\text{ - }}\,\sqrt {{{\text{m}}^{\text{2}}}\,{\text{ - }}\,{{\text{n}}^{\text{2}}}} {\text{)}}$.

Ans: $\dfrac{{\dfrac{{{\text{a + b}}}}{{\text{2}}}}}{{\sqrt {{\text{ab}}} }}\,{\text{ = }}\,\dfrac{{\text{m}}}{{\text{n}}}$

$\dfrac{{{\text{a + b}}}}{{{\text{2}}\sqrt {{\text{ab}}} }}{\text{ = }}\dfrac{{\text{m}}}{{\text{n}}}$

By C and D,

$\dfrac{{{\text{a + b + 2}}\sqrt {{\text{ab}}} }}{{{\text{a + b - 2}}\sqrt {{\text{ab}}} }}\,{\text{ = }}\,\dfrac{{{\text{m}}\,{\text{ + }}\,{\text{n}}}}{{{\text{m}}\,{\text{ - }}\,{\text{n}}}}$

$\Rightarrow\, \dfrac{{{{{\text{(}}\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} {\text{)}}}^{\text{2}}}}}{{{{{\text{(}}\sqrt {\text{a}} \,{\text{ - }}\,\sqrt {\text{b}} {\text{)}}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{m}}\,{\text{ + }}\,{\text{n}}}}{{{\text{m}}\,{\text{ - }}\,{\text{n}}}}$

$\Rightarrow\, \dfrac{{\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} }}{{\sqrt {\text{a}} \,{\text{ - }}\,\sqrt {\text{b}} }}\,{\text{ = }}\,\dfrac{{\sqrt {{\text{m}}\,{\text{ + }}\,{\text{n}}} }}{{\sqrt {{\text{m}}\,{\text{ - }}\,{\text{n}}} }}$

By C and D,

$\Rightarrow\,\dfrac{{\sqrt {\text{a}} }}{{\sqrt {\text{b}} }}{\text{ = }}\dfrac{{\sqrt {{\text{m}}\,{\text{ + }}\,{\text{n}}} \,{\text{ + }}\,\sqrt {{\text{m}}\,{\text{ - }}\,{\text{n}}} }}{{\sqrt {{\text{m}}\,{\text{ + }}\,{\text{n}}} \,{\text{ - }}\,\sqrt {{\text{m}}\,{\text{ - }}\,{\text{n}}} }}$

Squaring on both sides, we get,

$\dfrac{{\text{a}}}{{\text{b}}}\,{\text{ = }}\,\dfrac{{{\text{m  +  n  +  m  -  n  + 2}}\sqrt {{{\text{m}}^{\text{2}}}\,{\text{ - }}\,{{\text{n}}^{\text{2}}}} }}{{{\text{m  +  n  +  m  -  n  -  2}}\sqrt {{{\text{m}}^{\text{2}}}\,{\text{ - }}\,{{\text{n}}^{\text{2}}}} }}$

$\Rightarrow\,\dfrac{{\text{a}}}{{\text{b}}}\,{\text{ = }}\,\dfrac{{{\text{m}}\,{\text{ + }}\sqrt {{{\text{m}}^{\text{2}}}\,{\text{ - }}\,{{\text{n}}^{\text{2}}}} \,}}{{{\text{m}}\,{\text{ - }}\sqrt {{{\text{m}}^{\text{2}}}\,{\text{ - }}\,{{\text{n}}^{\text{2}}}} \,}} $

4. Between 1 and 31, m number have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of m.

Ans: Let 1, A1, A2, …., Am, 31 are in A.P.

a = 1, an = 31

am+2 = 314

${\text{an}}\,{\text{ = }}\,{\text{a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}$

${\text{31}}\,{\text{ = }}\,{\text{a}}\,{\text{ + }}\,{\text{(m}}\,{\text{ + }}\,{\text{2}}\,{\text{ - }}\,{\text{1)d}}$

${\text{d}}\,{\text{ = }}\,\dfrac{{{\text{30}}}}{{{\text{m}}\,{\text{ + }}\,{\text{1}}}}$

$\dfrac{{{\text{A7}}}}{{{\text{Am - 1}}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{9}}}{\text{(given)}}$

$\Rightarrow\, \dfrac{{{\text{1 + }}\,{\text{7}}\left( {\dfrac{{{\text{30}}}}{{{\text{m}}\,{\text{ + }}\,{\text{1}}}}} \right)}}{{{\text{1}}\,{\text{ + }}\,{\text{(m}}\,{\text{ - }}\,{\text{1)}}\left( {\dfrac{{{\text{30}}}}{{{\text{m}}\,{\text{ + }}\,{\text{1}}}}} \right)}}\,{\text{ = }}\,\dfrac{{\text{5}}}{{\text{9}}}$

$\Rightarrow\, {\text{m}}\,{\text{ = }}\,{\text{1}}$

5. The sum of two no. is 6 times their geometric mean, show that no. are in the ratio $(3\, + \,3\sqrt 2 )\,:\,(3\, - \,2\sqrt 2 )$.

Ans: ${\text{a  +  b  =  6}}\sqrt {{\text{ab}}}$

$\dfrac{{{\text{a  +  b}}}}{{2\sqrt {{\text{ab}}} }} = \dfrac{3}{1}$ 

By C and D, we get,

$\dfrac{{{\text{a  +  b  +  2}}\sqrt {{\text{ab}}} }}{{{\text{a  +  b  -  2}}\sqrt {{\text{ab}}} }}\, = \,\dfrac{{3 + 1}}{{3 - 1}}$

$\Rightarrow\, \dfrac{{{{{\text{(}}\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} {\text{)}}}^{\text{2}}}}}{{{{{\text{(}}\sqrt {\text{a}} \,{\text{ - }}\,\sqrt {\text{b}} {\text{)}}}^{\text{2}}}}}\,{\text{ = }}\,\dfrac{{\text{2}}}{{\text{1}}}$

$\Rightarrow\, \dfrac{{\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} }}{{\sqrt {\text{a}} \,{\text{ - }}\,\sqrt {\text{b}} }}\,{\text{ = }}\,\dfrac{{\sqrt 2 }}{{\text{1}}}$

Again by C and D, we get,

$\dfrac{{\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} \,{\text{ + }}\,\sqrt {\text{a}} \,{\text{ - }}\sqrt {\text{b}} }}{{\sqrt {\text{a}} \,{\text{ + }}\,\sqrt {\text{b}} \, - \,\sqrt {\text{a}} \, + \sqrt {\text{b}} }}\, = \,\dfrac{{\sqrt 2 \, + \,1}}{{\sqrt 2 \, - \,1}}$

$\Rightarrow\, \dfrac{{2\sqrt {\text{a}} }}{{2\sqrt {\text{b}} }}\, = \,\dfrac{{\sqrt 2 \, + \,1}}{{\sqrt 2 \, - \,1}}$

$\Rightarrow\, \dfrac{{\text{a}}}{{\text{b}}}\,{\text{ = }}\,\dfrac{{{{{\text{(}}\sqrt {\text{2}} \,{\text{ + }}\,{\text{1)}}}^{\text{2}}}}}{{{{{\text{(}}\sqrt {\text{2}} \,{\text{ - }}\,{\text{1)}}}^{\text{2}}}}}\,{\text{(Squaring}}\,{\text{both}}\,{\text{sides)}}$

$\Rightarrow\, \dfrac{{\text{a}}}{{\text{b}}}\,{\text{ = }}\,\dfrac{{2\, + \,1\, + \,2\sqrt {\text{2}} }}{{2\, + \,1\, - \,2\sqrt {\text{2}} }}$

$\Rightarrow\, \dfrac{{\text{a}}}{{\text{b}}}\,{\text{ = }}\,\dfrac{{3\, + \,2\sqrt {\text{2}} }}{{3\, - \,2\sqrt {\text{2}} }}$

$\Rightarrow\, {\text{a}}\,{\text{:}}\,{\text{b}}\,{\text{ = }}\,{\text{(3}}\,{\text{ + }}\,{\text{2}}\sqrt {\text{2}} {\text{)}}\,{\text{:}}\,{\text{(3}}\,{\text{ - }}\,{\text{2}}\sqrt {\text{2}} {\text{)}}$

Practice Question for Chapter 8- Sequence and Series

Some practice questions of chapter 8 - Sequence and Series are as follows.

1. Calculate the 8th term in the series 2, 6, 18, 54, ……

Answer: 4374.

2. Calculate the missing term in the series,  4, 12, 36, _, 324, 972.

Answer: 108

3. Calculate the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer: 3050

4. The ratio of the sums of the n terms of two arithmetic progressions is 5n+4 to 9n+6.  Calculate the ratio of their 18th terms.

Answer: 179:321

5. Find the missing term in the series, 2,7,9,3,8,11,4,9,13,_,10,15.

Answer: 5

Important Topics Covered in the Chapter: Sequence and Series  

Some important topics covered in Sequence and Series are as follows.

• Introduction to Sequences and Series

• Arithmetic Progression

• Geometric Progression

• Some Special Series’ Formulas of the Sum of n Terms

• Arithmetic Mean and Geometric Mean

Benefits of Important Questions for Class 11 Maths Chapter 8 - Sequence and Series

1. Focused Preparation: The PDF compiles critical questions, helping students concentrate on key concepts and problem types likely to appear in exams.

2. Comprehensive Coverage: Includes a variety of question types, from basic to advanced, ensuring thorough practice and better understanding.

3. Exam-Oriented: Curated by experts, the questions align with the latest syllabus and exam patterns, increasing the likelihood of performing well.

4. Time Management: Practising these important questions helps students learn to solve problems efficiently, improving time management during exams.

5. Step-by-Step Solutions: Detailed solutions provided in the PDF enable students to learn proper problem-solving techniques and clarify doubts instantly.

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Conclusion

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