Chapter 9 Sequence and Series

Sequence is a collection of objects taken one by one in which repetitions may present but order matters. It can have any number of terms. When these terms are added it is called a series. We get some general expressions to solve sequences and series. It solves very tedious calculations which are very difficult to solve manually. This chapter consists of arithmetic progression and mean, geometric progression and mean, general terms, sum of n terms, arithmetic and geometric series, infinite G.P, relation between A.M and G.M, some special series.

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Exercise 1

Exercise 2

  • Q1 Find the sum of odd integers from 1 to 2001.
    Ans:
    The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
     
    This sequence forms an A.P.
     
    Here, first term, a = 1
     
    Common difference, d = 2
     
    Here,
     
    \begin{align} a + (n - 1)d = 2001 \end{align}
     
    \begin{align} => 1 + (n - 1)(2) = 2001 \end{align}
     
    \begin{align} => 2n -2 = 2000 \end{align}
     
    \begin{align} => n = 1001 \end{align}
     
    \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}
     
    \begin{align} \therefore S_n = \frac {1001}{2}\left[2 × 1 + (1001 -1)×2\right]\end{align}
     
    \begin{align} = \frac {1001}{2}\left[2 + 1000×2\right]\end{align}
     
    \begin{align} = \frac {1001}{2} × 2002\end{align}
     
    \begin{align} =1001 × 1001 \end{align}
     
    \begin{align} = 1002001 \end{align}
     
    Thus, the sum of odd numbers from 1 to 2001 is 1002001.
     

    Q2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
    Ans:

    The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

    This sequence forms an A.P.
     
    Here, first term, a = 105
     
    Common difference, d = 5
     
    Here,
     
    \begin{align} a + (n - 1)d = 995 \end{align}
     
    \begin{align} => 105 + (n - 1)5 = 995 \end{align}
     
    \begin{align} =>  (n - 1)5 = 995 - 105 = 890 \end{align}
     
    \begin{align} => n -1 = 178 \end{align}
     
    \begin{align} => n = 179 \end{align}
     
    \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}
     
    \begin{align} \therefore S_n = \frac {179}{2}\left[2 × (105) + (179 -1)×(5)\right]\end{align}
     
    \begin{align} = \frac {179}{2}\left[2(105) + (178)(5)\right]\end{align}
     
    \begin{align} = 179\left[105 + (89)5\right]\end{align}
     
    \begin{align} = (179)\left[105 + 445\right]\end{align}
     
    \begin{align} =179 × 550 \end{align}
     
    \begin{align} = 98450 \end{align}
     

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

     


    Q3 In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
    Ans:

    First term = 2

    Let d be the common difference of the A.P.

    Therefore, 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

    According to the given condition,

    \begin{align} => 10 + 10d = \frac{1}{4}(10 + 35d) \end{align}

    \begin{align} => 40 + 40d = 10 + 35d \end{align}

    \begin{align} => 30 = - 5d \end{align}

    \begin{align} => d = - 6 \end{align}

    \begin{align} \therefore a_{20} = a + (20 -1)d = 2 + (19)(-6) = 2 - 114 = -112\end{align}

    Thus, the 20th term of the A.P. is –112.

     

     


    Q4 How many terms of the A.P. \begin{align} -6, -\frac{11}{2}, -5, ... \end{align} are needed to give the sum –25?
    Ans:

    Let the sum of n terms of the given A.P. be –25.

    It is known that, \begin{align} S_n = \frac {n}{2}\left[2a + (n -1)d\right]\end{align}, where n = number of terms, a = first term, and d = common difference

    Here, a = –6

    \begin{align} d = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2}\end{align}

    Therefore, we obtain

    \begin{align} -25 = \frac {n}{2}\left[2 × (-6) + (n -1)×\frac{1}{2}\right]\end{align}

    \begin{align} => -50 = n\left[-12  + \frac{n}{2} - \frac{1}{2}\right]\end{align}
     
    \begin{align} => -50 = n\left[-\frac{25}{2} + \frac{n}{2}\right]\end{align}
     
    \begin{align} => -100 = n\left(-25 + n\right)\end{align}
     
    \begin{align} => n^2 - 25n + 100 = 0\end{align}
     
    \begin{align} => n^2 - 5n -20n + 100 = 0\end{align}
     
    \begin{align} => n(n - 5)- 20(n - 5) = 0\end{align}
     
    \begin{align} => n = 20 \; or\; 5\end{align}

     


    Q5 In an A.P., if pth term is \begin{align} \frac{1}{q} \; and \;qth\; term \; is\; \frac{1}{p}\end{align} , prove that the sum of first pq terms is \begin{align} \frac{1}{2}(pq+1) \; where \; p ≠ q \end{align}
    Ans:

    It is known that the general term of an A.P. is an = a + (n – 1)d

    ∴ According to the given information,

    \begin{align}p^{th} \; term= a_p=a+(p-1)d=\frac{1}{q}  \;\; ...(1)\end{align}

    \begin{align}q^{th} \; term= a_q=a+(q-1)d=\frac{1}{p}  \;\; ...(2)\end{align}

    Subtracting (2) from (1), we obtain

    \begin{align} (p-1)d - (q-1)d=\frac{1}{q}-\frac{1}{p} \end{align}

    \begin{align} ⇒(p-1-q+1)d = \frac{p-q}{pq} \end{align}

    \begin{align} ⇒(p-q)d = \frac{p-q}{pq} \end{align}

    \begin{align} ⇒d = \frac{1}{pq} \end{align}

    Putting the value of d in (1), we obtain

    \begin{align} a + (p-1)\frac{1}{pq}= \frac{1}{q}\end{align}

    \begin{align} ⇒a = \frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq} \end{align}

    \begin{align} \therefore S_{pq} = \frac{pq}{2}\left[2a + (pq-1)d\right]\end{align}

    \begin{align}  = \frac{pq}{2}\left[\frac{2}{pq} + (pq-1)\frac{1}{pq}\right]\end{align}

    \begin{align}  = 1 + \frac{1}{2}(pq-1)\end{align}

    \begin{align}  =\frac{1}{2}pq + 1 - \frac{1}{2}=\frac{1}{2}pq + \frac{1}{2}\end{align}

    \begin{align}  =\frac{1}{2}(pq+1) \end{align}

    Thus, the sum of first pq terms of the A.P. is \begin{align}  =\frac{1}{2}(pq+1). \end{align}

     


    Q6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
    Ans:

    Let the sum of n terms of the given A.P. be 116.

    \begin{align}  S_n=\frac{n}{2}\left[2a + (n-1)d\right] \end{align}

    Here, a = 25 and d = 22 – 25 = – 3

    \begin{align}  \therefore S_n=\frac{n}{2}\left[2 × 25 + (n-1)(-3)\right] \end{align}

    \begin{align}  ⇒ 116=\frac{n}{2}\left[50 -3n +3\right] \end{align}

    \begin{align}  ⇒ 232=n(53-3n)=53n -3n^2 \end{align}

    \begin{align}  ⇒ 3n^2 -53n + 232 =0\end{align}

    \begin{align}  ⇒ 3n^2 -24n -29n + 232 =0\end{align}

    \begin{align}  ⇒ 3n(n-8) -29(n-8) =0\end{align}

    \begin{align}  ⇒ (n-8)(3n-29) =0\end{align}

    \begin{align}  ⇒ n=8 \;or\;n=\frac{29}{3} \end{align}

    However, n cannot be equal to \begin{align}  \frac{29}{3}. \end{align} Therefore, n = 8

    \begin{align}  \therefore a_8 = Last \; Term = a + (n-1)d= 25 + (8-1)(-3) \end{align}

    \begin{align}  = 25 + (7)(-3)=25-21 \end{align}

    \begin{align}  = 4 \end{align}

    Thus, the last term of the A.P. is 4.

     


    Q7 Find the sum to n terms of the A.P., whose kth term is 5k + 1.
    Ans:

    It is given that the kth term of the A.P. is 5k + 1.

    kth term = ak = + (k – 1)d

    ∴ + (k – 1)d = 5k + 1

    a + kd – d = 5k + 1

    Comparing the coefficient of k, we obtain d = 5

    – = 1

    ⇒ a – 5 = 1

    ⇒ a = 6

    \begin{align}   S_n = \frac{n}{2}\left[2a + (n-1)d\right] \end{align}

    \begin{align}   = \frac{n}{2}\left[2(6) + (n-1)(5)\right] \end{align}

    \begin{align}   = \frac{n}{2}\left[12 + 5n -5\right] \end{align}

    \begin{align}   = \frac{n}{2}\left(5n + 7\right) \end{align}

     


    Q8 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference
    Ans:

    It is known that,

    \begin{align}   S_n= \frac{n}{2}\left[2a + (n-1)d\right] \end{align}

    According to the given condition,

    \begin{align}    \frac{n}{2}\left[2a + (n-1)d\right] = pn + qn^2 \end{align}

    \begin{align}   ⇒ \frac{n}{2}\left[2a + nd-d\right] = pn + qn^2 \end{align}

    \begin{align}   ⇒ na + n^2\frac{d}{2} - n.\frac{d}{2}= pn + qn^2 \end{align}

    Comparing the coefficients of n2 on both sides, we obtain

    \begin{align}   \frac{d}{2} = q \end{align}

    \begin{align}  \therefore d = 2q \end{align}

    Thus, the common difference of the A.P. is 2q.

     


    Q9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
    Ans:

    Let a1a2, and d1d2 be the first terms and the common difference of the first and second arithmetic progression respectively.

    According to the given condition,

     

    \begin{align}  \frac{Sum \;of \;n \;terms \;of \;first\; A.P.}{Sum\; of \;n\; terms \;of \;second \;A.P.} = \frac{5n+4}{9n+6} \end{align}

    \begin{align}  ⇒\frac{\frac{n}{2}\left[2a_1 + (n-1)d_1\right]}{\frac{n}{2}\left[2a_2 + (n-1)d_2\right]} = \frac{5n+4}{9n+6} \end{align}

    \begin{align}  ⇒\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{5n+4}{9n+6} \;\;\;\;...(1)\end{align}

    Substituting n = 35 in (1), we obtain

    \begin{align}  ⇒\frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{5(35)+4}{9(35)+6} \end{align}

    \begin{align}  ⇒\frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{179}{321} \;\;\;\;...(2)\end{align}

    \begin{align}  \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{a_1 + 17d_1}{a_2 + 17d_2}  \;\;\;\;...(3)\end{align}

    From (2) and (3), we obtain

    \begin{align}  \frac{18^{th} \;term \;of\; first\; A.P.}{18^{th} \;term \;of\; second\; A.P.}=\frac{179}{321}\end{align}

    Thus, the ratio of 18th term of both the A.P.s is 179: 321.

     


    Q10 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
    Ans:

    Let a and d be the first term and the common difference of the A.P. respectively.

    Here,

    \begin{align} S_p = \frac{p}{2}\left[2a+(p-1)d\right]\end{align}

    \begin{align} S_q = \frac{q}{2}\left[2a+(q-1)d\right]\end{align}

    According to the given condition,

    \begin{align} \frac{p}{2}\left[2a+(p-1)d\right]=\frac{q}{2}\left[2a+(q-1)d\right]\end{align}

    \begin{align} ⇒p\left[2a+(p-1)d\right]=q\left[2a+(q-1)d\right]\end{align}

    \begin{align} ⇒2ap + pd(p-1)=2aq+qd(q-1)\end{align}

    \begin{align} ⇒2a(p-q) +d[p(p-1)-q(q-1)]=0\end{align}

    \begin{align} ⇒2a(p-q) +d[p^2 -p -q^2 +q]=0\end{align}

    \begin{align} ⇒2a(p-q) +d[(p-q)(p+q)-(p-q)]=0\end{align}

    \begin{align} ⇒2a(p-q) +d[(p-q)(p+q-1)]=0\end{align}

    \begin{align} ⇒2a +d(p+q-1)=0\end{align}

    \begin{align} ⇒d=\frac{-2a}{p+q-1} \;\;\;\;...(1)\end{align}

    \begin{align} \therefore S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).d\right] \end{align}

    \begin{align} ⇒ S_{p+q}=\frac{p+q}{2}\left[2a+(p+q-1).\left(\frac{-2a}{p+q-1}\right)\right] \;\;\;\; [From (1)] \end{align}

    \begin{align}=\frac{p+q}{2}\left[2a-2a\right]\end{align}

    \begin{align}=0\end{align}

    Thus, the sum of the first (p + q) terms of the A.P. is 0.

     


    Q11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that 
    Ans:

    Let a1 and d be the first term and the common difference of the A.P. respectively.

    According to the given information,

    S subscript p space end subscript equals space p over 2 open square brackets 2 a subscript 1 plus space open parentheses p minus 1 close parentheses d close square brackets equals a
rightwards double arrow 2 a subscript 1 space plus space open parentheses p minus 1 close parentheses d space equals space fraction numerator 2 a over denominator p end fraction space space space space space space... left parenthesis 1 right parenthesis
S subscript q equals space q over 2 open square brackets 2 a subscript 1 plus space open parentheses q minus 1 close parentheses d close square brackets equals b
rightwards double arrow 2 a subscript 1 space plus space open parentheses q minus 1 close parentheses d space equals space fraction numerator 2 b over denominator q end fraction space space space space space space... left parenthesis 2 right parenthesis
S subscript r equals space r over 2 open square brackets 2 a subscript 1 plus space open parentheses r minus 1 close parentheses d close square brackets equals c
rightwards double arrow 2 a subscript 1 space plus space open parentheses r minus 1 close parentheses d space equals space fraction numerator 2 c over denominator r end fraction space space space space space space... left parenthesis 3 right parenthesis
S u b t r a c t i n g space left parenthesis 2 right parenthesis space f r o m space left parenthesis 1 right parenthesis comma space w e space o b t a i n
left parenthesis p minus 1 right parenthesis d space minus left parenthesis q minus 1 right parenthesis d equals fraction numerator 2 a over denominator p end fraction equals fraction numerator 2 b over denominator q end fraction
rightwards double arrow d open parentheses p minus 1 minus q plus 1 close parentheses equals fraction numerator 2 a q minus 2 b q over denominator p q end fraction
rightwards double arrow d left parenthesis p minus q right parenthesis equals fraction numerator 2 a q minus 2 b p over denominator p q end fraction
rightwards double arrow d equals fraction numerator 2 open parentheses a q minus b p close parentheses over denominator p q open parentheses p minus q close parentheses end fraction space space space space space space space space space space space... left parenthesis 4 right parenthesis
S u b t r a c t i n g space left parenthesis 3 right parenthesis space f r o m space left parenthesis 2 right parenthesis comma space w e space o b t a i n
open parentheses q minus 1 close parentheses d minus open parentheses r minus 1 close parentheses d equals fraction numerator 2 b over denominator q end fraction minus fraction numerator 2 c over denominator r end fraction
rightwards double arrow d open parentheses q minus 1 minus r plus 1 close parentheses equals fraction numerator 2 b over denominator q end fraction minus fraction numerator 2 c over denominator r end fraction
rightwards double arrow d open parentheses q minus r close parentheses equals fraction numerator 2 b r minus 2 q c over denominator q r end fraction
rightwards double arrow d equals fraction numerator 2 open parentheses b r minus q c close parentheses over denominator q r open parentheses q minus r close parentheses end fraction space space space space space space space space space space space... left parenthesis 5 right parenthesis
E q u a t i n g space b o t h space t h e space v a l u e space o f space d space o b t a i n e d space i n space left parenthesis 4 right parenthesis space a n d space left parenthesis 5 right parenthesis comma space w e space o b t a i n
fraction numerator a q minus b p over denominator p q open parentheses p minus q close parentheses end fraction equals fraction numerator b r minus q c over denominator q r open parentheses q minus r close parentheses end fraction
rightwards double arrow q r open parentheses q minus r close parentheses open parentheses a q minus b q close parentheses equals p q open parentheses p minus q close parentheses open parentheses b r minus q c close parentheses
rightwards double arrow r open parentheses a q minus b p close parentheses open parentheses q minus r close parentheses equals p open parentheses b r minus q c close parentheses open parentheses p minus q close parentheses
rightwards double arrow open parentheses a q r minus b p r close parentheses open parentheses q minus r close parentheses equals open parentheses b p r minus p q c close parentheses open parentheses p minus q close parentheses
D i v i d i n g space b o t h space s i d e s space b space p q r ; w e space o b a t i n
open parentheses a over p minus b over q close parentheses open parentheses q minus r close parentheses equals open parentheses b over q minus c over r close parentheses open parentheses p minus q close parentheses
rightwards double arrow a over p open parentheses q minus r close parentheses minus b over q open parentheses q minus r plus p minus q close parentheses plus c over r open parentheses p minus q close parentheses equals 0
rightwards double arrow a over p open parentheses q minus r close parentheses plus b over q open parentheses r minus p close parentheses plus c over r open parentheses p minus q close parentheses equals 0
T h u s comma t h e space g i v e n space r e s u l t space i s space p r o v e d.


    Q12 The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).
    Ans:

    Let a and b be the first term and the common difference of the A.P. respectively.

    According to the given condition,

    fraction numerator S u m space o f space m space t e r m s over denominator s u m space o f space n space t e r m s end fraction equals m squared over n squared
rightwards double arrow fraction numerator begin display style m over 2 end style open square brackets 2 a plus open parentheses m minus 1 close parentheses d close square brackets over denominator begin display style n over 2 end style open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets end fraction equals m squared over n squared
rightwards double arrow space space space fraction numerator 2 a plus left parenthesis m minus 1 right parenthesis d over denominator 2 a plus left parenthesis n minus 1 right parenthesis d end fraction equals m over n space space space space space.... left parenthesis 1 right parenthesis
P u t t i n g space m space equals space 2 m space  space 1 space a n d space n space equals space 2 n space  space 1 space i n space left parenthesis 1 right parenthesis comma space w e space o b t a i n
fraction numerator 2 a plus left parenthesis 2 m minus 2 right parenthesis d over denominator 2 a plus left parenthesis 2 n minus 2 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction
rightwards double arrow fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction space space space space.... left parenthesis 2 right parenthesis
fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator a plus left parenthesis m minus 1 right parenthesis d over denominator a plus left parenthesis n minus 1 right parenthesis d end fraction space space space space... left parenthesis 3 right parenthesis
F r o m space left parenthesis 2 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space o b t a i n
fraction numerator m to the power of t h end exponent t e r m space o f space A. P. over denominator n to the power of t h space end exponent t e r m space o f space A. P. end fraction equals fraction numerator 2 m minus 1 over denominator 2 n minus 1 end fraction
T h u s comma space t h e space g i v e n space r e s u l t space i s space p r o v e d.


    Q13 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
    Ans:

    Let a and b be the first term and the common difference of the A.P. respectively.

    am = a + (m – 1)d = 164 … (1)

    Sum of n terms,S subscript n space space end subscript equals space n over 2 open square brackets 2 a plus left parenthesis n minus 1 right parenthesis d close square brackets

    Here,

    n over 2 open square brackets 2 a plus n d minus d close square brackets equals 3 n squared plus 5 n
rightwards double arrow n a space plus space n squared. d over 2 equals 3 n squared plus 5 n

    Comparing the coefficient of n2 on both sides, we obtain

    d over 2 equals 3
rightwards double arrow d equals 6

    Comparing the coefficient of n on both sides, we obtain

    a minus d over 2 equals 5
rightwards double arrow a minus 3 equals 5
rightwards double arrow a equals 8

    Therefore, from (1), we obtain

    8 + (m – 1) 6 = 164

    ⇒ (m – 1) 6 = 164 – 8 = 156

    ⇒ – 1 = 26

    ⇒ m = 27

    Thus, the value of m is 27.


    Q14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
    Ans:

    Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that

    8, A1, A2, A3, A4, A5, 26 is an A.P.

    Here, = 8, = 26, n = 7

    Therefore, 26 = 8 + (7 – 1) d

    ⇒ 6d = 26 – 8 = 18

    ⇒ = 3

    A1 = a + d = 8 + 3 = 11

    A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

    A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

    A4 = a + 4= 8 + 4 × 3 = 8 + 12 = 20

    A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

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


    Q15 If is the A.M. between a and b, then find the value of n.
    Ans:

    A.M. of a and b = fraction numerator a plus b over denominator 2 end fraction

    According to the given condition,

    fraction numerator a plus b over denominator 2 end fraction equals fraction numerator a to the power of n space space space end exponent plus space b to the power of n over denominator a to the power of n minus 1 end exponent plus b to the power of n minus 1 end exponent end fraction
rightwards double arrow open parentheses a plus b close parentheses open parentheses a to the power of n minus 1 end exponent plus b to the power of n minus 1 end exponent close parentheses equals 2 open parentheses a to the power of n plus b to the power of n close parentheses
rightwards double arrow a to the power of n plus a b to the power of n minus 1 end exponent plus b a to the power of n minus 1 end exponent plus b to the power of n equals 2 a to the power of n plus 2 b to the power of n
rightwards double arrow a b to the power of n minus 1 end exponent plus a to the power of n minus 1 end exponent b equals a to the power of n plus b to the power of n
rightwards double arrow a b to the power of n minus 1 end exponent minus b to the power of n equals a to the power of n minus a to the power of n minus 1 end exponent b
rightwards double arrow b to the power of n minus 1 end exponent open parentheses a minus b close parentheses equals a to the power of n minus 1 end exponent open parentheses a minus b close parentheses
rightwards double arrow b to the power of n minus 1 end exponent equals a to the power of n minus 1 end exponent
rightwards double arrow open parentheses a over b close parentheses to the power of n minus 1 end exponent equals 1 equals open parentheses a over b close parentheses to the power of 0
rightwards double arrow n minus 1 equals 0
rightwards double arrow n equals 1

     


    Q16 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)thnumbers is 5:9. Find the value of m.
    Ans:

    Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.

    Here, a = 1, b = 31, n = m + 2

    ∴ 31 = 1 + (m + 2 – 1) (d)

    ⇒ 30 = (m + 1) d

    rightwards double arrow d equals fraction numerator 30 over denominator m plus 1 end fraction space space space space space space space space... left parenthesis 1 right parenthesis

    A1 = a + d

    A2 = a + 2d

    A3 = a + 3d …

    ∴ A7 = a + 7d

    Am–1 = a + (m – 1) d

    According to the given condition,

    fraction numerator a plus 7 d over denominator a plus open parentheses m minus 1 close parentheses d end fraction equals 5 over 9

rightwards double arrow fraction numerator 1 plus 7 open parentheses begin display style fraction numerator 30 over denominator open parentheses m plus 1 close parentheses end fraction end style close parentheses over denominator 1 plus open parentheses m minus 1 close parentheses open parentheses begin display style fraction numerator 30 over denominator m plus 1 end fraction end style close parentheses end fraction equals 5 over 9 space space space space space space space space space space space space space space space space open square brackets F r o m space left parenthesis 1 right parenthesis close square brackets

rightwards double arrow fraction numerator m plus 1 plus 7 open parentheses 30 close parentheses over denominator m plus 1 plus 30 open parentheses m minus 1 close parentheses end fraction equals 5 over 9

rightwards double arrow fraction numerator m plus 1 plus 210 over denominator m plus 1 plus 30 m minus 30 end fraction equals 5 over 9

rightwards double arrow fraction numerator m plus 211 over denominator 31 m space minus 29 end fraction equals 5 over 9

rightwards double arrow 9 m space plus space 1899 space equals space 155 m space minus 145

rightwards double arrow 155 m space minus 9 m space equals space 1899 space plus 145

rightwards double arrow 146 m equals 2044
rightwards double arrow m equals 14

    Thus, the value of m is 14.


    Q17 A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?
    Ans:

    The first installment of the loan is Rs 100.

    The second installment of the loan is Rs 105 and so on.

    The amount that the man repays every month forms an A.P.

    The A.P. is 100, 105, 110, …

    First term, a = 100

    Common difference, d = 5

    A30 = a + (30 – 1)d

    = 100 + (29) (5)

    = 100 + 145

    = 245

    Thus, the amount to be paid in the 30th installment is Rs 245.


    Q18 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
    Ans:

    The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.

    It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).

    So

      S subscript n equals 180 to the power of 0 open parentheses n minus 2 close parentheses
rightwards double arrow n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets equals 180 to the power of 0 open parentheses n minus 2 close parentheses
rightwards double arrow n over 2 open square brackets 240 to the power of 0 plus open parentheses n minus 1 close parentheses 5 to the power of 0 close square brackets equals 180 open parentheses n minus 2 close parentheses
rightwards double arrow n open square brackets 240 plus open parentheses n minus 1 close parentheses 5 close square brackets equals 360 open parentheses n minus 2 close parentheses
rightwards double arrow 240 n space plus space 5 n squared space minus 5 n space equals space 360 n space minus 720
rightwards double arrow 5 n squared plus 235 n minus 360 n space plus 720 equals 0
rightwards double arrow 5 n squared space minus 125 n space plus 720 equals 0
rightwards double arrow n squared minus 25 n space plus space 144 space equals space 0
rightwards double arrow n squared minus 16 n minus 9 n plus 144 equals 0
rightwards double arrow n open parentheses n minus 16 close parentheses minus 9 open parentheses n minus 16 close parentheses equals 0
rightwards double arrow open parentheses n minus 9 close parentheses open parentheses n minus 16 close parentheses equals 0
rightwards double arrow n equals 9 space o r space 16


Exercise 3

Exercise 4

Exercise 5

  • Q1 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
    Ans:

    Let a and d be the first term and the common difference of the A.P. respectively.

    It is known that the kth term of an A. P. is given by

    ak = a + (k –1) d

    ∴ am + n = a + (m + n –1) d

    am – n = a + (m – n –1) d

    am = a + (m –1) d

    ∴ am + n + am – n = a + (m + n –1) d + a + (m – n –1) d

    = 2a + (m + n –1 + m – n –1) d

    = 2a + (2m – 2) d

    = 2a + 2 (m – 1) d

    =2 [a + (m – 1) d]

    = 2am

    Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.


    Q2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
    Ans:

    Let the three numbers in A.P. be a – da, and a + d.

    According to the given information,

    (a – d) + (a) + (a + d) = 24 … (1)

    ⇒ 3a = 24

    ∴ a = 8

    (a – da (a + d) = 440 … (2)

    ⇒ (8 – d) (8) (8 + d) = 440

    ⇒ (8 – d) (8 + d) = 55

    ⇒ 64 – d2 = 55

    ⇒ d2 = 64 – 55 = 9

    ⇒ = ± 3

    Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.

    Thus, the three numbers are 5, 8, and 11.


    Q3 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)
    Ans:

    Let a and b be the first term and the common difference of the A.P. respectively.

    Therefore,

    S subscript 1 space equals space n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
S subscript 2 space equals space fraction numerator 2 n over denominator 2 end fraction open square brackets 2 a plus open parentheses 2 n minus 1 close parentheses d close square brackets space equals space n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space space... left parenthesis 2 right parenthesis
S subscript 3 space equals space fraction numerator 3 n over denominator 2 end fraction open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets space space space space space... left parenthesis 3 right parenthesis

    From (1) and (2), we obtain

    S subscript 2 space minus space S subscript 1 space equals n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space minus n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets
space space space space space space space space space space space space space space space space equals space n open curly brackets fraction numerator 4 a space plus space 4 n d space minus 2 d space minus 2 a space minus n d space plus d over denominator 2 end fraction close curly brackets
space space space space space space space space space space space space space space space space equals space n open square brackets fraction numerator 2 a space plus space 3 n d space minus d over denominator 2 end fraction close square brackets
space space space space space space space space space space space space space space space equals space n over 2 space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets
therefore space 3 open parentheses S subscript 2 space minus space S subscript 1 close parentheses space equals space fraction numerator 3 n over denominator 2 end fraction space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets equals S subscript 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets F r o m space left parenthesis 3 right parenthesis close square brackets

    Hence, the given result is proved.


    Q4 Find the sum of all numbers between 200 and 400 which are divisible by 7.
    Ans:

    The numbers lying between 200 and 400, which are divisible by 7, are

    203, 210, 217, ­­­­­­­­… 399

    ∴First term, a = 203

    Last term, l = 399

    Common difference, d = 7

    Let the number of terms of the A.P. be n.

    ∴ an = 399 = a + (n –1) d

    ⇒ 399 = 203 + (n –1) 7

    ⇒ 7 (n –1) = 196

    ⇒ n –1 = 28

    ⇒ n = 29

    therefore space S subscript 29 space equals space 29 over 2 space open parentheses 203 space plus space 399 close parentheses
space space space space space space space space space space space space space equals space 29 over 2 space open parentheses 602 close parentheses
space space space space space space space space space space space space space equals space open parentheses 29 close parentheses open parentheses 301 close parentheses
space space space space space space space space space space space space space equals space 8729

    Thus, the required sum is 8729.


    Q5 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
    Ans:

    The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

    This forms an A.P. with both the first term and common difference equal to 2.

    ⇒100 = 2 + (n –1) 2

    ⇒ n = 50

    therefore space 2 plus 4 plus 6 plus... plus 100 space equals 50 over 2 space open square brackets 2 open parentheses 2 close parentheses space plus space open parentheses 50 minus 1 close parentheses open parentheses 2 close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 over 2 space open square brackets 4 space plus space 98 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 25 close parentheses open parentheses 102 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2550

    The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

    This forms an A.P. with both the first term and common difference equal to 5.

    ∴100 = 5 + (n –1) 5

    ⇒ 5n = 100

    ⇒ n = 20

    therefore space 5 space plus space 10 space plus space... space plus space 100 space equals space 20 over 2 space open square brackets 2 open parentheses 5 close parentheses space plus space open parentheses 20 space minus 1 close parentheses 5 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space open parentheses 19 close parentheses 5 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 10 space open square brackets 10 space plus space 95 close square brackets space equals space 10 space cross times space 105
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1050

    The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

    This also forms an A.P. with both the first term and common difference equal to 10.

    ∴100 = 10 + (n –1) (10)

    ⇒ 100 = 10n

    ⇒ n = 10

    therefore space 10 space plus space 20 space plus space... space plus space 100 space equals space 10 over 2 space open square brackets 2 open parentheses 10 close parentheses space plus space open parentheses 10 space minus 1 close parentheses open parentheses 10 close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 5 space open square brackets 20 space plus space 90 close square brackets space equals space 5 open parentheses 110 close parentheses space equals space 550

    ∴Required sum = 2550 + 1050 – 550 = 3050

    Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.


    Q6 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
    Ans:

    The two-digit numbers, which when divided by 4, yield 1 as remainder, are

    13, 17, … 97.

    This series forms an A.P. with first term 13 and common difference 4.

    Let n be the number of terms of the A.P.

    It is known that the nth term of an A.P. is given by, an = a + (n –1) d

    ∴97 = 13 + (n –1) (4)

    ⇒ 4 (n –1) = 84

    ⇒ n – 1 = 21

    ⇒ n = 22

    Sum of n terms of an A.P. is given by,

    S subscript n space equals space n over 2 space open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets
therefore space S subscript 22 space equals space 22 over 2 space open square brackets 22 open parentheses 13 close parentheses space plus space open parentheses 22 minus 1 close parentheses open parentheses 4 close parentheses close square brackets
space space space space space space space space space space space space space equals space 11 open square brackets 26 space plus space 84 close square brackets
space space space space space space space space space space space space space equals space 1210

    Thus, the required sum is 1210.


    Q7 If f is a function satisfying f(x +y) = f(x) f(y) for all x,y N  such that f(1) = 3  and  , find the value of n.
    Ans:

    It is given that,

    (x + y) = (x) × (y) for all xy ∈ N … (1)

    (1) = 3

    Taking x = y = 1 in (1), we obtain

    f (1 + 1) = (2) = (1) (1) = 3 × 3 = 9

    Similarly,

    (1 + 1 + 1) = (3) = (1 + 2) = (1) (2) = 3 × 9 = 27

    (4) = (1 + 3) = f (1) (3) = 3 × 27 = 81

    ∴ (1), (2), (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.

    It is known that, S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus space 1 close parentheses over denominator r space minus 1 end fraction

    It is given that, sum from x equals 1 to n of space f left parenthesis x right parenthesis space equals space 120

    therefore space 120 space equals space fraction numerator 3 open parentheses 3 to the power of n space minus 1 close parentheses over denominator 3 minus 1 end fraction
rightwards double arrow 120 space equals space 3 over 2 open parentheses 3 to the power of n space minus space 1 close parentheses
rightwards double arrow 3 to the power of n space minus space 1 space equals space 80
rightwards double arrow 3 to the power of n space equals space 81 space equals space 3 to the power of 4 space
therefore space n space equals 4

    Thus, the value of n is 4.


    Q8 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
    Ans:

    Let the sum of n terms of the G.P. be 315.

    It is known that, S subscript n space equals space fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction

    It is given that the first term a is 5 and common ratio is 2.

    therefore space 315 space equals fraction numerator 5 open parentheses 2 to the power of n minus 1 close parentheses space over denominator 2 minus 1 end fraction
rightwards double arrow 2 to the power of n space end exponent minus 1 space equals space 63
rightwards double arrow 2 to the power of n space equals 64 space equals space open parentheses 2 close parentheses to the power of 6
rightwards double arrow n equals 6

    ∴Last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160

    Thus, the last term of the G.P. is 160.


    Q9 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 a and r be the first term and the common ratio of the G.P. respectively.

    ∴ a = 1

    a3 = ar2 = r2

    a5 = ar4 = r4

    ∴ r2 + r4 = 90

    ⇒ r4 + r2 – 90 = 0

    rightwards double arrow r squared space equals space fraction numerator minus 1 space plus space square root of 1 plus space 360 end root over denominator 2 end fraction equals fraction numerator minus 1 space plus-or-minus square root of 361 over denominator 2 end fraction equals fraction numerator minus 1 plus-or-minus 19 over denominator 2 end fraction equals minus 10 space o r space 9

    ∴ r = ± 3                                  ( Taking real roots)

    Thus, the common ratio of the G.P. is ±3.


    Q10 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
    Ans:

    Let the three numbers in G.P. be aar, and ar2.

    From the given condition, a + ar + ar2 = 56

    ⇒ a (1 + r + r2) = 56

    rightwards double arrow a space equals fraction numerator 56 over denominator 1 plus r plus r squared end fraction space space space space... space left parenthesis 1 right parenthesis

    a – 1, ar – 7, ar2 – 21 forms an A.P.

    ∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

    ⇒ ar – a – 6 = ar2 – ar – 14

    ar2 – 2ar + a = 8

    ar2 – ar – ar + a = 8

    a(r2 + 1 – 2r) = 8

    ⇒ (r – 1)2 = 8 … (2)

    rightwards double arrow fraction numerator 56 over denominator 1 plus r plus r squared end fraction open parentheses r minus 1 close parentheses squared equals 8 space space space space space space space space space space open square brackets U sin g space left parenthesis 1 right parenthesis close square brackets

    ⇒7(r2 – 2r + 1) = 1 + r + r2

    ⇒7r2 – 14 r + 7 – 1 – r – r2 = 0

    ⇒ 6r2 – 15r + 6 = 0

    ⇒ 6r2 – 12r – 3r + 6 = 0

    ⇒ 6r (r – 2) – 3 (r – 2) = 0

    ⇒ (6r – 3) (r – 2) = 0

    therefore space r space equals space 2 comma 1 half

    When r = 2, a = 8

    When r space equals space 1 half space comma space a space equals space 32

    Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

    When r space equals space 1 half, the three numbers in G.P. are 32, 16, and 8.

    Thus, in either case, the three required numbers are 8, 16, and 32.


    Q11 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
    Ans:

    Let the G.P. be T1, T2, T3, T4, … T2n.

    Number of terms = 2n

    According to the given condition,

    T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]

    ⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0

    ⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1]

    Let the G.P. be aarar2ar3, …

    therefore space fraction numerator a r space open parentheses r to the power of n space minus 1 close parentheses over denominator r minus 1 end fraction space equals space fraction numerator 4 space cross times space a open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction
rightwards double arrow a r space equals space 4 a
rightwards double arrow r space equals space 4

    Thus, the common ratio of the G.P. is 4.


    Q12 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
    Ans:

    Let the A.P. be aa + da + 2da + 3d, ... a + (n – 2) da + (n – 1)d.

    Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

    Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] = 4a + (4n – 10) d

    According to the given condition,

    4a + 6d = 56

    ⇒ 4(11) + 6d = 56 [Since a = 11 (given)]

    ⇒ 6d = 12

    ⇒ d = 2

    ∴ 4a + (4n –10) d = 112

    ⇒ 4(11) + (4n – 10)2 = 112

    ⇒ (4n – 10)2 = 68

    ⇒ 4n – 10 = 34

    ⇒ 4n = 44

    ⇒ n = 11

    Thus, the number of terms of the A.P. is 11.


    Q13 If 
    Ans:

    It is given that,

    fraction numerator a space plus space b x over denominator a space minus b x end fraction space equals space fraction numerator b space plus space c x over denominator b space minus c x end fraction
rightwards double arrow space open parentheses a space plus space b x close parentheses open parentheses b minus c x close parentheses space equals space open parentheses b plus c x close parentheses open parentheses a minus b x close parentheses
rightwards double arrow a b space minus space a c x space plus space b squared x space minus space b c x squared space equals space a b space minus space b squared x space plus space a c x space minus space b c x squared
rightwards double arrow 2 b squared x space equals space 2 a c x
rightwards double arrow b squared space equals space a c
rightwards double arrow b over a space equals space c over b space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
A l s o comma space fraction numerator b plus c x over denominator b minus c x end fraction space equals space fraction numerator c space plus d x over denominator c space minus d x end fraction space
rightwards double arrow open parentheses b plus c x close parentheses open parentheses c minus d x close parentheses space equals space open parentheses b minus c x close parentheses open parentheses c plus d x close parentheses
rightwards double arrow b c space minus space b d x space plus space c squared x space minus space c d x squared space equals space b c space plus space b d x space minus space c squared x space minus space c d x squared
rightwards double arrow 2 c squared x space equals space 2 b d x
rightwards double arrow c squared space equals space b d
rightwards double arrow c over d space equals space d over c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

    From (1) and (2), we obtain

    b over a space equals space c over b space equals d over c

    Thus, abc, and d are in G.P.


    Q14 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn
    Ans:

    Let the G.P. be aarar2ar3, … arn – 1

    According to the given information,

    S space equals space fraction numerator a space open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction
P space equals space a to the power of n space end exponent cross times space r to the power of 1 plus 2 plus... plus n minus 1 end exponent
space space space space equals space a to the power of n space end exponent r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction space space space space space space space open square brackets because space S u m space o f space f i r s t space n space n a t u r a l space n u m b e r s space i s space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets end exponent

R space equals space 1 over a space plus space fraction numerator 1 over denominator a r end fraction space plus space... space plus space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction
space space space space equals space fraction numerator r to the power of n minus 1 end exponent plus r to the power of n minus 2 end exponent space plus space... r plus 1 over denominator a r to the power of n minus 1 end exponent end fraction
space space space space equals space fraction numerator 1 open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction space cross times space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction space space space space space space space space space space open square brackets because space 1 comma r comma... r to the power of n minus 1 end exponent space f o r m s space a space G. P. close square brackets
space space space space equals space fraction numerator r to the power of n space minus 1 over denominator a r to the power of n minus 1 end exponent open parentheses r minus 1 close parentheses end fraction
therefore space P squared R to the power of n space equals space a to the power of 2 n end exponent r to the power of n open parentheses n minus 1 close parentheses space end exponent fraction numerator open parentheses r to the power of n space minus 1 close parentheses to the power of n over denominator a to the power of n r to the power of n open parentheses n minus 1 close parentheses end exponent open parentheses r minus 1 close parentheses to the power of n end fraction
space space space space space space space space space space space space space space space space space equals space fraction numerator a to the power of n open parentheses r to the power of n minus 1 close parentheses to the power of n over denominator open parentheses r minus 1 close parentheses to the power of n end fraction
space space space space space space space space space space space space space space space space space equals space open square brackets fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction close square brackets to the power of n
space space space space space space space space space space space space space space space space equals space S to the power of n

    Hence P2 Rn= Sn