Important Questions for CBSE Class 12 Maths Chapter 8 - Sequences and Series 2024-25

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

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