NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series - Exercise 9.2

Exercise 9.2 

1. Find the sum of odd integers from 1 to 2001.

Ans: In between 1 and 2001, the odd integers are 1, 3, 5,…, 1999, 2001.

The given sequence forms an A.P. with its first term, a = 1 and common difference, d = 2

Here, 

$a+(n-1)d=2001$ 

   $\Rightarrow {\text{1}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)(2)}}\,{\text{ = }}\,{\text{2001}}$ 

  $ \Rightarrow \,{\text{2n}}\,{\text{ - }}\,{\text{2}}\,{\text{ = 2001}} $

$   \Rightarrow \,{\text{n}}\,{\text{ = }}\,{\text{1001}}$ 

  ${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$ 

 $ \therefore \,{\text{Sn}}\,{\text{ = }}\,\dfrac{1001}{2}[2\times 1+(1001-1)\times 2]$

${\text{Sn}}\,{\text{ = }}\,\dfrac{1001}{2}[2+1000\times 2]$

${\text{Sn}}\,{\text{ = }}\, 1001\times 1001 $

${\text{ = }}\,{\text{1002001}}$

Therefore, the sum of odd numbers between 1 and 2001 is 1002001.

2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Ans: The multiples of 5 lying between 100 and 1000 are 105, 110,…, 995.

This sequence forms an A.P. with its first term, a = 105 and common difference, d = 5

${\text{a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}\, = \,995$ 

  $ \Rightarrow {\text{105}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)(5)}}\,{\text{ = }}\,995$ 

  $ \Rightarrow \,({\text{n}}\,{\text{ - }}\,{\text{1}})5\,{\text{ = 995}}\,{\text{ - }}\,{\text{105}}\,{\text{ = }}\,{\text{890}} $

  $ \Rightarrow \,{\text{n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,178 $

   $\Rightarrow \,{\text{n}}\,{\text{ = }}\,{\text{179}}$   

$\therefore \,{\text{Sn}}\,{\text{ = }}\,\dfrac{{{\text{179}}}}{{\text{2}}}\left[ {{\text{2(105)}}\,{\text{ + }}\,{\text{(179}}\,{\text{ - }}\,{\text{1)(5)}}} \right]$

$   = \,\dfrac{{179}}{2}\left[ {2(105)\, + \,(178)(5)} \right]$ 

$= 179\left[ {105\, + \,(89)5} \right] $

 $  = 179(105\, + \,445) $

   = 179(550) 

   = 98450  

Therefore, the sum of natural numbers lying between 100 and 1000, which are multiples of 5 are 98450.

3. In an A.P., the first term is 2 and the sum of the first five term is one fourth of the next five terms. Show that ${20^{{\text{th}}}}$ term is -112.

Ans: Given: a = 2

Let the common difference be d.

Thus, the A.P. is 2, 2+d, 2+2d, 2+3d, …

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the given condition,

$10+10d=\dfrac{1}{4}(10+35d)$

$   \Rightarrow {\text{40}}\,{\text{ + }}\,{\text{40d}}\,{\text{ =  10  + }}\,{\text{35d}} $

  $ \Rightarrow {\text{ 30  =   - 5d}} $

$   \Rightarrow {\text{ d  =   - 6}} $

$  \therefore {\text{ a20  =  a  +  (20  -  1)d  =  2  +  19( - 6)  =  2  -  114  =   - 112}}$

Therefore, ${20^{{\text{th}}}}$ term of the given A.P. is -112.

4. How many terms of the A.P. $ - 6,\,\dfrac{{ - 11}}{2},\, - 5,\,...$are needed to give the sum -25?

Ans: Let the sum of n terms of the A.P. be -25.

We know that,

${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$

Here, n = number of terms, a = first term, d = common difference.

Here, a = -6

d = $\dfrac{{ - 11}}{2} + 6\, = \,\dfrac{{ - 11 + 12}}{2}\, = \,\dfrac{1}{2}$

Thus, we get,

$ {\text{ - 25  =  }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2( - 6)  +  (n  -  1)}}\left( {\dfrac{1}{2}} \right)} \right] $

   $\Rightarrow {\text{  - 50  =  n}}\left[ { - 12\, + \,\dfrac{{\text{n}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right] $

   $\Rightarrow \, - 50\, = \,{\text{n}}\left[ {\dfrac{{{\text{ - 25}}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{n}}}{{\text{2}}}} \right]$

$   \Rightarrow \, - 100\, = \,{\text{n( - 25}}\,{\text{ + }}\,{\text{n)}} $

 $  \Rightarrow \,{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{25n}}\,{\text{ + }}\,{\text{100}}\,{\text{ = }}\,{\text{0}} $

   $\Rightarrow \,{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{5n}}\,{\text{ - }}\,{\text{20n}}\,{\text{ + 100}}\,{\text{ = }}\,{\text{0}} $

  $ \Rightarrow {\text{n(n}}\,{\text{ - }}\,{\text{5)}}\,{\text{ - 20(n}}\,{\text{ - }}\,{\text{5)}}\,{\text{ = }}\,{\text{0}} $

$   \Rightarrow {\text{n}}\,{\text{ = }}\,{\text{20}}\,{\text{or}}\,{\text{5}} $

5. In an A.P., if the ${{\text{p}}^{{\text{th}}}}$term is 1/q and ${{\text{q}}^{{\text{th}}}}$term is 1/p, prove that the sum of first pq terms is

$\dfrac{{\text{1}}}{{\text{2}}}\left( {{\text{pq}}\,{\text{ + }}\,{\text{1}}} \right)$, where ${\text{p}}\, \ne \,{\text{q}}$.

Ans: We know that, the general form of A.P. is ${\text{an}}\,{\text{ = }}\,{\text{a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}$

From the given data,

${{\text{p}}^{{\text{th}}}}{\text{ term  =  ap  =  a  +  (p  -  1)d  =  }}\dfrac{1}{{\text{q}}}\,\,......(1) $

  ${{\text{q}}^{{\text{th}}}}{\text{ term  =  aq  =  a  +  (q  -  1)d  =  }}\dfrac{{\text{1}}}{{\text{p}}}\,\,........(2) $

Subtracting (2) from (1), we get,

$ ({\text{p  -  1}}){\text{d  -  (q  -  1)d  =  }}\dfrac{{\text{1}}}{{\text{q}}}\,{\text{ - }}\,\dfrac{{\text{1}}}{{\text{p}}} $

 $  \Rightarrow \,{\text{(p  -  1  -  q  +  1)d  =  }}\dfrac{{{\text{p}}\,{\text{ - }}\,{\text{q}}}}{{{\text{pq}}}} $

  $ \Rightarrow {\text{ (p  -  q)d  =  }}\dfrac{{{\text{p}}\,{\text{ - }}\,{\text{q}}}}{{{\text{pq}}}} $

  $ \Rightarrow {\text{ d  =  }}\dfrac{{\text{1}}}{{{\text{pq}}}} $

Substituting the value of d in (1), we get,

${\text{a  +  (p  -  1)}}\dfrac{{\text{1}}}{{{\text{pq}}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{q}}}$

$ \Rightarrow {\text{a  =  }}\dfrac{{\text{1}}}{{\text{q}}}\,{\text{ - }}\,\dfrac{{\text{1}}}{{\text{q}}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{{\text{pq}}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{{\text{pq}}}}$

$\therefore \,{\text{Spq  =  }}\dfrac{{{\text{pq}}}}{{\text{2}}}\left[ {{\text{2a  +  (pq  - 1)d}}} \right]$

  ${\text{ =  }}\dfrac{{{\text{pq}}}}{{\text{2}}}\left[ {\dfrac{{\text{2}}}{{{\text{pq}}}}\,{\text{ + }}\,{\text{(pq}}\,{\text{ - }}\,{\text{1)}}\dfrac{{\text{1}}}{{{\text{pq}}}}} \right] $

  ${\text{ = }}\,{\text{1}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}}{\text{(pq}}\,{\text{ - }}\,{\text{1)}} $

  ${\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}{\text{pq}}\,{\text{ + }}\,{\text{1}}\,{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\,{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}{\text{pq}}\,{\text{ + }}\,\dfrac{{\text{1}}}{{\text{2}}} $

$  {\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}{\text{(pq}}\,{\text{ + }}\,{\text{1)}} $

Therefore, sum of the first pq terms is $\dfrac{{\text{1}}}{{\text{2}}}{\text{(pq}}\,{\text{ + }}\,{\text{1)}}$.

6. If the sum of a certain number of terms in an A.P. 25, 22, 19, … is 116. Find the last term.

Ans: Let the sum of n terms of an A.P. be 116.

${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$

Here, a = 25 and d = 25 – 22 = -3

$  \therefore {\text{Sn  =  }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2 x 25}}\,{\text{ + }}\,{\text{(n - 1)( - 3)}}} \right]$

 $  \Rightarrow {\text{ 116  =  }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{50}}\,{\text{ - }}\,{\text{3n}}\,{\text{ + }}\,{\text{3}}} \right] $

  $ \Rightarrow \,{\text{232}}\,{\text{ = }}\,{\text{n(53}}\,{\text{ - }}\,{\text{3n)}}\,{\text{ = }}\,{\text{53n}}\,{\text{ - }}\,{\text{3}}{{\text{n}}^{\text{2}}} $

 $  \Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{53n}}\,{\text{ + 232}}\,{\text{ = }}\,{\text{0}} $

  $ \Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{24n}}\,{\text{ - }}\,{\text{29n}}\,{\text{ + 232}}\,{\text{ = }}\,{\text{0}} $

 $  \Rightarrow \,{\text{3n(n - 8)}}\,{\text{ - }}\,{\text{29(n - 8)}}\,{\text{ = }}\,{\text{0}} $

   $\Rightarrow \,{\text{(n - 8)(3n}}\,{\text{ - }}\,{\text{29)}}\,{\text{ = }}\,{\text{0}} $

   $\Rightarrow \,{\text{n}}\,{\text{ = }}\,{\text{8}}\,{\text{or}}\,{\text{n}}\,{\text{ = }}\dfrac{{{\text{29}}}}{{\text{3}}} $  

We know that, n cannot be equal to $\dfrac{{29}}{3}$, therefore n = 8.

$ \therefore {a_8}{\text{ = }}\,{\text{Last term  =  a  +  (n  -  1)d  =  25  +  (8  -  1)( - 3) }} $

${\text{ =   25  +  (7)( - 3)  =  25  -  21}} $

 $ {\text{ =  4}} $ 

Therefore, the last term of the given A.P. is 4.

7. Find the sum to n terms of an A.P., whose ${{\text{K}}^{{\text{th}}}}$term is 5k+1.

Ans: Given: ${{\text{K}}^{{\text{th}}}}$ term is 5k+1

$ {{\text{K}}^{{\text{th}}}}{\text{ term  =  a  +  (k  -  1)d}} $

  $\therefore {\text{ a  +  (k  -  1)d  =  5k  +  1}} $

  ${\text{a  +  kd  -  d  =  5k  +  1}} $ 

By comparing the coefficient of k, we get d = 5.

$\Rightarrow {\text{a  -  d  =  1}} $

   $\Rightarrow {\text{a  -  5  =  1}} $

   $\Rightarrow {\text{a  =  6}} $

 $ {\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right] $

 $ {\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2(6)}}\,{\text{ + }}\,{\text{(n - 1)(5)}}} \right] $

  ${\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{12}}\,{\text{ + }}\,{\text{5n}}\,{\text{ - }}\,{\text{5}}} \right] $

  ${\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{5n}}\,{\text{ + }}\,{\text{7}}} \right] $ 

8. If the sum of n terms of an A.P. is $({\text{pn  +  q}}{{\text{n}}^{\text{2}}})$, where p and q are constants. Find the common difference.

Ans: We know that, 

${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$

From the given data,

$\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a  +  (n  -  1)d}}} \right]{\text{ =  pn  +  q}}{{\text{n}}^{\text{2}}} $

   $\Rightarrow \dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a  +  nd  -  d}}} \right] = {\text{pn  +  q}}{{\text{n}}^{\text{2}}} $

  $ \Rightarrow {\text{na  +  }}{{\text{n}}^{\text{2}}}\dfrac{{\text{d}}}{{\text{2}}}\,{\text{ - }}\,{\text{n}}{\text{.}}\dfrac{{\text{d}}}{{\text{2}}}\,{\text{ = }}\,{\text{pn  +  q}}{{\text{n}}^{\text{2}}} $ 

 By comparing the coefficients of ${{\text{n}}^{\text{2}}}$ on both the sides,

$\dfrac{{\text{d}}}{{\text{2}}}\,{\text{ = }}\,{\text{q}} $

  $\therefore {\text{d  =  2q}} $ 

Therefore, common difference d = 2q.

9. The sums of n terms of the arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their ${18^{{\text{th}}}}$terms.

Ans: Let ${\text{a1,}}\,{\text{a2}}$ be the first terms and ${\text{d1,}}\,{\text{d2}}$ be the common differences of first and second arithmetic progression respectively.

From the given condition,

\[\Rightarrow \dfrac{\dfrac{\text{n}}{\text{2}}\left[ \text{2}{{\text{a}}_{\text{1}}}\text{+}\left( \text{n-1} \right){{\text{d}}_{\text{1}}} \right]}{\dfrac{\text{n}}{\text{2}}\left[ \text{2}{{\text{a}}_{\text{2}}}\text{+}\left( \text{n-1} \right){{\text{d}}_{\text{2}}} \right]}\text{=}\dfrac{\text{5n+4}}{\text{9n+6}}\] …..(1)

Substitute n = 35 in (1)

\[\Rightarrow \dfrac{\text{2}{{\text{a}}_{\text{1}}}\text{+34}{{\text{d}}_{\text{1}}}}{\text{2}{{\text{a}}_{\text{2}}}\text{+34}{{\text{d}}_{\text{2}}}}\text{=}\dfrac{\text{5}\left( \text{35} \right)\text{+4}}{\text{9}\left( \text{35} \right)\text{+6}}\]

\[\Rightarrow \dfrac{{{\text{a}}_{\text{1}}}\text{+17}{{\text{d}}_{\text{1}}}}{{{\text{a}}_{\text{2}}}\text{+17}{{\text{d}}_{\text{2}}}}\text{=}\dfrac{\text{179}}{\text{321}}\] …….(2)

\[\dfrac{\text{18th}\,\,\text{term}\,\,\text{of}\,\text{first}}{\text{18th term of second A}\text{.P}}\text{=}\]\[\dfrac{{{\text{a}}_{\text{1}}}\text{+17}{{\text{d}}_{\text{1}}}}{{{\text{a}}_{\text{2}}}\text{+17}{{\text{d}}_{\text{2}}}}\]……(3)

From (2) and (3) we get.

\[\dfrac{\text{18th}\,\,\text{term}\,\,\text{of}\,\text{first}}{\text{18th term of second A}\text{.P}}\text{=}\dfrac{179}{321}\]

Therefore ratio of both 18th terms of the A.P are 179:321

10. If the sum of first p terms of an A.P. is equal to the sum of first q terms, then find the sum of first (p+q) terms.

Ans: Let a, d be the first term and common difference of the given term respectively.

$ {\text{Sp}}\,{\text{ = }}\,\dfrac{{\text{p}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(p}}\,{\text{ - }}\,{\text{1)d}}} \right] $

$  {\text{Sq}}\,{\text{ = }}\,\dfrac{{\text{q}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(q}}\,{\text{ - }}\,{\text{1)d}}} \right] $ 

 From the given condition,

$\dfrac{{\text{p}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(p}}\,{\text{ - }}\,{\text{1)d}}} \right]\, = \,\dfrac{{\text{q}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(q}}\,{\text{ - }}\,{\text{1)d}}} \right] $

  $ \Rightarrow {\text{ p}}\left[ {{\text{2a  +  (p  -  1)d}}} \right] = {\text{q}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(q}}\,{\text{ - }}\,{\text{1)d}}} \right] $ 

$ \Rightarrow {\text{ 2ap  +  pd(p  -  1)  =  2aq  +  qd(q  -  1)}} $

   $\Rightarrow {\text{ 2a(p  -  q)  +  d}}\left[ {{\text{p(p  -  1)  -  q(q  -  1)}}} \right]\, = \,0 $

  $ \Rightarrow {\text{ 2a(p  -  q)  +  d}}\left[ {{{\text{p}}^{\text{2}}}{\text{  -  p  - }}{{\text{q}}^{\text{2}}}{\text{  +  q}}} \right]\, = \,0 $

  $ \Rightarrow {\text{ 2a(p  -  q)  +  d}}\left[ {{\text{(p  -  q)(p  +  q) - (p  -  q)}}} \right]\, = \,0 $

  $ \Rightarrow {\text{ 2a(p  -  q)  +  d}}\left[ {{\text{(p  -  q)(p  +  q  -  1)}}} \right]\, = \,0 $

  $ \Rightarrow {\text{ 2a  +  d(p  +  q  -  1)  =  0}} $

  $ \Rightarrow {\text{ d  =  }}\dfrac{{{\text{ - 2a}}}}{{{\text{p  +  q  - 1}}}}\,.....(1) $  

$  \therefore {\text{Sp + q}}\,{\text{ = }}\,\dfrac{{{\text{p + q}}}}{2}\left[ {{\text{2a  +  (p  +  q  -  1)}}{\text{.d}}} \right] $

  $ \Rightarrow {\text{ Sp + q}}\,{\text{ = }}\,\dfrac{{{\text{p + q}}}}{2}\left[ {{\text{2a  +  (p  +  q  -  1)}}{\text{.}}\left( {\dfrac{{{\text{ - 2a}}}}{{{\text{p  +  q  - 1}}}}} \right)} \right] $

  $ = \,\dfrac{{{\text{p  +  q}}}}{2}\left[ {{\text{2a  -  2a}}} \right] $

   $= \,0 $  

Therefore, the sum of first (p+q) terms is 0.

11. Sum of the first p, q and r terms of an A.P. are a, b and c respectively. Prove that 

$\dfrac{{\text{a}}}{{\text{p}}}{\text{(q}}\,{\text{ - }}\,{\text{r)}}\,{\text{ + }}\,\dfrac{{\text{b}}}{{\text{q}}}{\text{(r}}\,{\text{ - }}\,{\text{p)}}\,{\text{ + }}\,\dfrac{{\text{c}}}{{\text{r}}}{\text{(p}}\,{\text{ - }}\,{\text{q)}}\,{\text{ = }}\,{\text{0}}$

Ans: Let the first term and common difference be ${\text{a1}}\ ,{\text{and}}\,{\text{d}}$ respectively.

From the given data,

$ {\text{Sp}}\,{\text{ = }}\,\dfrac{{\text{p}}}{{\text{2}}}\left[ {{\text{2a1  +  (p  -  1)d}}} \right] = \,{\text{a}}\, $

   $\Rightarrow {\text{ 2a1  +  (p  -  1)d = }}\,\dfrac{{{\text{2a}}}}{{\text{p}}}\,.....{\text{(1)}} $

  ${\text{Sq}}\,{\text{ = }}\,\dfrac{{\text{q}}}{{\text{2}}}\left[ {{\text{2a1  +  (q  -  1)d}}} \right]{\text{ = }}\,{\text{b}}\, $

  $ \Rightarrow {\text{ 2a1  +  (q  -  1)d = }}\,\dfrac{{{\text{2b}}}}{{\text{q}}}\,.....{\text{(2)}} $

  ${\text{Sr}}\,{\text{ = }}\,\dfrac{{\text{r}}}{{\text{2}}}\left[ {{\text{2a1  +  (r  -  1)d}}} \right]{\text{ = }}\,{\text{c}}\, $

  $ \Rightarrow {\text{ 2a1  +  (r  -  1)d = }}\,\dfrac{{{\text{2c}}}}{{\text{r}}}\,.....{\text{(3)}} $  

Subtracting (2) from (1), we get,

${\text{(p  -  1)d  -  (q  -  1)d  =  }}\dfrac{{{\text{2a}}}}{{\text{p}}}\,{\text{ - }}\,\dfrac{{{\text{2b}}}}{{\text{q}}} $

 $  \Rightarrow \,{\text{d(p  -  1  -  q  +  1)  =  }}\dfrac{{{\text{2aq}}\,{\text{ - }}\,{\text{2bp}}}}{{{\text{pq}}}} $

  $ \Rightarrow \,{\text{d(p}}\,{\text{ - }}\,{\text{q)}}\,{\text{ = }}\,\dfrac{{{\text{2aq}}\,{\text{ - }}\,{\text{2bp}}}}{{{\text{pq}}}} $

   $\Rightarrow \,{\text{d}}\,{\text{ = }}\,\dfrac{{{\text{2(aq}}\,{\text{ - }}\,{\text{bp)}}}}{{{\text{pq(p}}\,{\text{ - }}\,{\text{q)}}}}\,....(4) $ 

Subtracting (3) from (2), we get,

$  {\text{(q  -  1)d  -  (r  -  1)d  =  }}\dfrac{{{\text{2b}}}}{{\text{q}}}\,{\text{ - }}\,\dfrac{{{\text{2c}}}}{{\text{r}}} $

  $ \Rightarrow \,{\text{d(q  -  1  -  r  +  1)  =  }}\dfrac{{{\text{2b}}}}{{\text{q}}}\,{\text{ - }}\,\dfrac{{{\text{2c}}}}{{\text{r}}} $

   $\Rightarrow {\text{d(q  -  r)  =  }}\dfrac{{{\text{2br - 2cq}}}}{{{\text{qr}}}} $

$   \Rightarrow {\text{d  =  }}\dfrac{{{\text{2(br - cq)}}}}{{{\text{qr(q  -  r)}}}}{\text{ }}.....{\text{(5)}} $  

Equating both values of d obtained from (4) and (5), we get,

$\dfrac{{{\text{aq  -  bp}}}}{{{\text{pq(p  -  q)}}}}\, = \,\dfrac{{{\text{br  -  qc}}}}{{{\text{qr(q  -  r)}}}} $

   $\Rightarrow {\text{ qr(q  -  r)(aq  -  bp)  =  pq(p  -  q)(br  -  qc)}} $

   $\Rightarrow \,{\text{r(aq  -  bp)(q  -  r)  =  p(br  -  qc)(p  -  q)}} $

 $  \Rightarrow {\text{ (aqr  -  bpr)(q  -  r)  =  (bpr  -  pqc)(p  -  q)}} $ 

Divide both sides by pqr, we get,

$ \left( {\dfrac{{\text{a}}}{{\text{p}}}{\text{ - }}\dfrac{{\text{b}}}{{\text{q}}}} \right){\text{(q}}\,{\text{ - }}\,{\text{r)}}\,{\text{ = }}\,\left( {\dfrac{{\text{b}}}{{\text{q}}}{\text{ - }}\dfrac{{\text{c}}}{{\text{r}}}} \right){\text{(p}}\,{\text{ - }}\,{\text{q)}} $

  $ \Rightarrow \,\dfrac{{\text{a}}}{{\text{p}}}{\text{(q}}\,{\text{ - }}\,{\text{r)}}\,{\text{ - }}\,\dfrac{{\text{b}}}{{\text{q}}}{\text{(q}}\,{\text{ - }}\,{\text{r}}\,{\text{ + }}\,{\text{p}}\,{\text{ - }}\,{\text{q)}}\,{\text{ + }}\,\dfrac{{\text{c}}}{{\text{r}}}{\text{(p}}\,{\text{ - }}\,{\text{q)}}\,{\text{ = }}\,{\text{0}} $

 $  \Rightarrow \dfrac{{\text{a}}}{{\text{p}}}{\text{(q}}\,{\text{ - }}\,{\text{r)}}\,{\text{ + }}\,\dfrac{{\text{b}}}{{\text{q}}}{\text{(r}}\,{\text{ - }}\,{\text{p)}}\,{\text{ + }}\,\dfrac{{\text{c}}}{{\text{r}}}{\text{(p}}\,{\text{ - }}\,{\text{q)}}\,{\text{ = }}\,{\text{0}} $ 

Hence, the result is proved.

12. The ratio of the sums of m and n terms of an A.P. is ${{\text{m}}^{\text{2}}}\,{\text{:}}\,{{\text{n}}^{\text{2}}}$. Show that the ratio of ${{\text{m}}^{{\text{th}}}}\,{\text{and}}\,{{\text{n}}^{{\text{th}}}}$term is (2m -1) : (2n - 1).

Ans: Let the first term and common difference be a and d respectively.

From the given condition,

$\dfrac{{{\text{Sum of m terms}}}}{{{\text{Sum of n terms}}}}\, = \,\dfrac{{{{\text{m}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}} $

 $ \dfrac{{\dfrac{{\text{m}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(m}}\,{\text{ - }}\,{\text{1)d}}} \right]}}{{\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]}}\,{\text{ = }}\,\dfrac{{{{\text{m}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}} $

  $ \Rightarrow \dfrac{{{\text{2a  +  (m  -  1)d}}}}{{{\text{2a  +  (n  -  1)d}}}} = \,\dfrac{{\text{m}}}{{\text{n}}}\,.....(1) $ 

Put m = 2m – 1 and n = 2n – 1, we get,

$\dfrac{{{\text{2a  +  (2m  - 2)d}}}}{{{\text{2a  +  (2n  -  2)d}}}} = \dfrac{{{\text{2m  -  1}}}}{{{\text{2n  -  1}}}} $

   $\Rightarrow \dfrac{{{\text{a  +  (m  -  1)d}}}}{{{\text{a}}\, + ({\text{n  -  1}}){\text{d}}}} = \dfrac{{{\text{2m  -  1}}}}{{{\text{2n  -  1}}}}\,.....(2) $

$ \dfrac{{{{\text{m}}^{{\text{th}}}}{\text{ terms of an A}}{\text{.P}}{\text{.}}}}{{{{\text{n}}^{{\text{th}}}}{\text{ terms of an A}}{\text{.P}}{\text{.}}}}\, = \,\dfrac{{{\text{a  +  (m  -  1)d}}}}{{{\text{a}}\, + ({\text{n  -  1}}){\text{d}}}}\,....(3) $ 

From (2) and (3), we get,

$\dfrac{{{{\text{m}}^{{\text{th}}}}{\text{ terms of an A}}{\text{.P}}{\text{.}}}}{{{{\text{n}}^{{\text{th}}}}{\text{ terms of an A}}{\text{.P}}{\text{.}}}}\, = \,\dfrac{{{\text{2m  -  1}}}}{{{\text{2n  -  1}}}}$

Therefore, the result is proved.

13. If the sum of n terms of an A.P. of ${\text{3}}{{\text{n}}^{\text{2}}}\,{\text{ + }}\,{\text{5n}}$and its mth term is 164. Find the value of m.

Ans: Let the first term and common difference be a and d respectively.

${\text{am  =  a  +  (m  -  1)d  =  164 }}.....{\text{(1)}}$

Sum of n terms: ${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$

Given,

$ \dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]\, = \,{\text{3}}{{\text{n}}^{\text{2}}}\,{\text{ + }}\,{\text{5n}} $

 $ \Rightarrow \,{\text{na  +  }}{{\text{n}}^{\text{2}}}{\text{.}}\dfrac{{\text{d}}}{{\text{2}}}\,{\text{ - }}\,\dfrac{{{\text{nd}}}}{{\text{2}}}\,{\text{ = }}\,{\text{3}}{{\text{n}}^{\text{2}}}\,{\text{ + }}\,{\text{5n}} $ 

By comparing the coefficient on both the sides, we get,

$ \dfrac{{\text{d}}}{2}\, = \,3 $

 $  \Rightarrow {\text{d}}\,{\text{ = }}\,{\text{6}} $ 

By comparing the coefficient of n on both the sides,

$  {\text{a}}\,{\text{ - }}\,\dfrac{{\text{d}}}{{\text{2}}}\,{\text{ = }}\,{\text{5}} $

$   \Rightarrow \,{\text{a}}\,{\text{ - }}\,{\text{3}}\,{\text{ = }}\,{\text{5}} $

$   \Rightarrow {\text{a}}\,{\text{ = }}\,{\text{8}} $ 

Therefore from (1), we get,

${\text{8  +  (m  -  1)6  =  164}} $

  $ \Rightarrow {\text{ (m  -  1)6  =  164  -  8  =  156}} $

$   \Rightarrow {\text{ m  -  1  =  26}} $

$   \Rightarrow {\text{ m  =  27}} $ 

Therefore, the value of m is 27.

14. Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Ans: Let A1, A2, A3, A4 and A5 be the 5 numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P.

Here, a = 8, b = 26, n = 7.

Thus, 

$ {\text{26  =  8  +  (7  -  1)d}} $

 $  \Rightarrow {\text{6d  =  26  -  8  =  18}} $

  $ \Rightarrow {\text{d  =  3}} $

$  {\text{A1  =  a  +  d  =  8  +  3  =  11}} $

 $ {\text{A2  =  a  +  2d  =  8  +  2(3)  =  14}} $

$  {\text{A3  =  a  +  3d  =  8  +  3(3)  =  17}} $

$  {\text{A4  =  a  +  4d  =  8  +  4(3)  =  20}} $

$  {\text{A5  =  8  +  5d  =  8  +  5(3)  =  23}} $ 

Therefore, the five numbers between 8 and 26 are 11, 14, 17, 20, 23.

15. If $\dfrac{{{{\text{a}}^{\text{n}}}\,{\text{ + }}\,{{\text{b}}^{\text{n}}}}}{{{{\text{a}}^{{\text{n - 1}}}}\,{\text{ + }}\,{{\text{b}}^{{\text{n - 1}}}}}}$is the A.M. between a and b then, find the value of n.

Ans: We know that, A.M. of a and b is $\dfrac{{{\text{a + b}}}}{{\text{2}}}$

From the given condition,

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

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

$ \Rightarrow {\text{ }}{{\text{a}}^{\text{n}}}{\text{  +  a}}{{\text{b}}^{{\text{n - 1}}}}{\text{  +  b}}{{\text{a}}^{{\text{n - 1}}}}{\text{  +  }}{{\text{b}}^{\text{n}}}{\text{  =  2}}{{\text{a}}^{\text{n}}}\,{\text{ + }}\,2{{\text{b}}^{\text{n}}} $

$ \Rightarrow {\text{ a}}{{\text{b}}^{{\text{n - 1}}}}{\text{  +  b}}{{\text{a}}^{{\text{n - 1}}}}\, = \,{{\text{a}}^{\text{n}}}\,{\text{ + }}\,{{\text{b}}^{\text{n}}} $

$\Rightarrow \,{\text{a}}{{\text{b}}^{{\text{n - 1}}}}\, - \,{{\text{b}}^{\text{n}}}\, = \,{{\text{a}}^{\text{n}}}\, - \,{\text{b}}{{\text{a}}^{{\text{n - 1}}}} $

$\Rightarrow \,{{\text{b}}^{{\text{n - 1}}}}{\text{(a}}\,{\text{ - }}\,{\text{b)}}\, = \,{{\text{a}}^{{\text{n - 1}}}}{\text{(a}}\,{\text{ - }}\,{\text{b)}} $

$\Rightarrow \,{{\text{b}}^{{\text{n - 1}}}}\, = \,{{\text{a}}^{{\text{n - 1}}}} $

$ \Rightarrow {\left( {\dfrac{{\text{a}}}{{\text{b}}}} \right)^{{\text{n - 1}}}} = 1 = {\left( {\dfrac{{\text{a}}}{{\text{b}}}} \right)^{\text{0}}} $

$   \Rightarrow \,{\text{n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,{\text{0}} $

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

16. Between 1 and 31, m numbers 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 A1, A2, …., Am be the m numbers between 1 and 31 such that 1, A1, A2, …., Am, 31 are in A.P.

Here, a = 1, b = 31, n = m + 2

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

$   \Rightarrow 30 = {\text{ (m  +  1)d}} $

$   \Rightarrow {\text{d  =  }}\dfrac{{30}}{{{\text{m  +  1}}}}\,......(1) $

$  {\text{A1  =  a  +  d}} $

$  {\text{A2  =  a  +  2d}} $

$  {\text{A3  =  a  +  3d}} $

$  \therefore {\text{A7  =  a  +  7d}} $

$  {\text{Am - 1  =  a  +  (m  -  1)d}} $  

From the given condition,

$\dfrac{{{\text{a  +  7d}}}}{{{\text{a  +  (m  -  1)d}}}} = \dfrac{5}{9} $

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

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

  $ \Rightarrow {\text{ }}\dfrac{{{\text{m  +  1  +  210}}}}{{{\text{m  +  1  +  30m  - 30}}}} = \dfrac{5}{9} $

$   \Rightarrow \,\dfrac{{{\text{m  +  211}}}}{{{\text{31m  - 29}}}} = \dfrac{5}{9} $

$   \Rightarrow {\text{ 9m  +  1899  =  155m  -  145}} $

$   \Rightarrow \,{\text{155m  -  9m  =  1899  +  145}} $

$   \Rightarrow {\text{ 146m  =  2044}} $

$   \Rightarrow {\text{ m  =  14}} $ 

Therefore, the value of m is 14.

17. A man starts repaying a loan as the first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount will he pay in the 30th installment?

Ans: Given: First installment = Rs.100

Second installment = Rs. 105 and so on.

The amount paid by man forms an A.P.

The A.P. is 100, 105, 110, ….

Here, a = 100, d = 5

$A30\, = \,{\text{a  +  (30  -  1)d}} $

$  {\text{ =  100  +  (29)(5)}} $

$  {\text{ =  100  +  145}} $

$  {\text{ =  245}} $ 

Therefore, the amount to be paid on the 30th installment is Rs. 245.

18. The difference between any two consecutive interior angles of a polygon is 5o . If the smallest angle is 120o, find the number of sides of the polygon.

Ans: The polygon angles form an A.P. with common difference as 5o and first term a as 120o

We know that, sum of all angles of a polygon with n sides = 180(n - 2).

$\therefore {\text{Sn  =  18}}{{\text{0}}^ \circ }({\text{n - 2)}} $

 $  \Rightarrow {\text{ }}\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{2a  +  (n  -  1)d}}} \right]{\text{ = }}\,{\text{18}}{{\text{0}}^{\text{o}}}{\text{(n - 2)}} $

$   \Rightarrow \,\dfrac{{\text{n}}}{{\text{2}}}\left[ {{\text{240}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)5}}} \right]\,{\text{ = }}\,{\text{180(n}}\,{\text{ - }}\,{\text{2)}} $ 

$\Rightarrow \,{\text{n}}\left[ {{\text{240}}\,{\text{ + }}\,{\text{(n}}\,{\text{ - }}\,{\text{1)5}}} \right]\,{\text{ = }}\,{\text{360(n}}\,{\text{ - }}\,{\text{2)}} $

$   \Rightarrow \,2{\text{40n}}\,{\text{ + }}\,{\text{5}}{{\text{n}}^{\text{2}}}\,{\text{ - 5n}}\,{\text{ = }}\,{\text{360n}}\,{\text{ - }}\,{\text{720}} $

 $  \Rightarrow {\text{5}}{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{125n}}\,{\text{ + 720}}\,{\text{ = }}\,{\text{0}} $

$   \Rightarrow {{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{25n}}\,{\text{ + }}\,{\text{144}}\,{\text{ = }}\,{\text{0}} $

  $ \Rightarrow \,{{\text{n}}^{\text{2}}}\,{\text{ - }}\,{\text{16n}}\,{\text{ - }}\,{\text{9n}}\,{\text{ + }}\,{\text{144}}\,{\text{ = }}\,{\text{0}} $

   $\Rightarrow {\text{n(n}}\,{\text{ - }}\,{\text{16)}}\,{\text{ - }}\,{\text{9(n}}\,{\text{ - }}\,{\text{16)}}\,{\text{ = }}\,{\text{0}} $

$   \Rightarrow ({\text{n}}\,{\text{ - }}\,{\text{9)(n}}\,{\text{ - }}\,{\text{16)}}\,{\text{ = }}\,{\text{0}} $

 $  \Rightarrow \,{\text{n}}\,{\text{ = }}\,{\text{9}}\,{\text{or}}\,{\text{16}} $ 

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 9 Exercises

All Important Topics and Formulae to Solve Exercise 9.2 — Class 11 NCERT Maths 

Exercise 9.2 is based on the series of Arithmetics Progressions (AP) and related formulae. To solve the second exercise of the chapter Sequences and Series of Class 11 NCERT Mathematics, you must remember the below given formulae.

If the AP series is a1, a2, a3, …. an, …., and the difference between the consecutive terms is d (constant), then 

• The nth term of AP series an = a + (n – 1) d

• The series can be written as a, a + d, a + 2d, a + 3d, …..

• Total sum of n terms of AP, ${\text{Sn}}\,{\text{ = }}\,\dfrac{{\text{n}}}{{\text{2}}}\left[{{\text{2a}}\,{\text{+}}\,{\text{(n}}\,{\text{ - }}\,{\text{1)d}}} \right]$ or Sn = \[\frac{n}{2}\](a + l)Sn

where a = first digit of AP sequence, n = number of digits, l = last digit

Arithmetic Mean: Arithmetic mean is the average of two numbers. If the two numbers are x and y, then their arithmetic mean will be expressed as 

Arithmetic Mean = \[\frac{x + y}{2}\]

Arithmetic mean is calculated when we have given two numbers, and we have to make an AP series by inserting numbers between them.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.2

Opting for the NCERT solutions for Ex 9.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Sequences and Series textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 9 Exercise 9.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 9 Exercise 9.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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