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.

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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

In order to give the students a step-by-step introduction to Sequence and Series, Vedantu experts developed Important Questions of Sequence and Series Class 11. The NCERT curriculum was carefully followed in the creation of all the content and solutions for Sequence and Series Class 11 Important Questions, allowing the students to use the content to get ready for the test. 

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