NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

 

NCERT solutions for class 11 maths chapter 9 Sequences and Series: Sequence means the progression of numbers in a definite order and series means the sum of the objects of the sequence. In the previous classes, you have studied about arithmetic progression(A.P). In this chapter, we will discuss more arithmetic progression(A.P) and geometric progression(G.P). In this article, you will get NCERT solutions for class 11 maths chapter 9 sequences and series. Important topics like arithmetic progression(A.P), geometric progression(G.P), arithmetic means(A.M), geometric mean(G.M), the relationship between A.M. and G.M, sum to n terms of special series, sum to n terms of squares and cubes of natural numbers are covered in this chapter. You will get questions related to these topics in the solutions of NCERT for class 11 maths chapter 9 sequences and series. In this chapter, there are two types of sequence.

  • Finite sequence( A sequence containing a finite number of terms)
  • Infinite sequence( A sequence has a first term but doesn't have last term or a sequence which is not finite)

Check all NCERT solutions from class 6 to 12 to understand the concepts in a much easy way. There are four exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

Miscellaneous Exercise

Topics of NCERT Grade 11 Maths Chapter- 9 Sequences and Series

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.5 Geometric Progression (G.P.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n terms of Special Series

The complete Solutions of NCERT Class 11 Mathematics Chapter 9 is provided below:

 

NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.1

Question:1  Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

           a _n = n ( n +2)

Answer:

Given : a _n = n ( n +2)

a _1 = 1 ( 1 +2)=3

a _2 = 2 ( 2 +2)=8

a _3 = 3 ( 3 +2)=15

a _4 = 4 ( 4 +2)=24

a _5 = 5 ( 5 +2)=35

Therefore, the required number of terms =3, 8, 15, 24, 35

Question:2 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are: 

      a _n = \frac{n }{n+1}

Answer:

Given : a _n = \frac{n }{n+1}

a _1 = \frac{1}{1+1}=\frac{1}{2}

a _2 = \frac{2}{2+1}=\frac{2}{3}

a _3 = \frac{3}{3+1}=\frac{3}{4}

a _4 = \frac{4}{4+1}=\frac{4}{5}

a _5 = \frac{5}{5+1}=\frac{5}{6}

Therefore, the required number of terms \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}

Question:3 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are: 

    a _ n = 2 ^n

Answer:

Given : a _ n = 2 ^n

a _ 1 = 2 ^1=2

a _ 2 = 2 ^2=4

a _ 3 = 2 ^3=8

a _ 4 = 2 ^4=16

a _ 5 = 2 ^5=32

Therefore, required number of terms =2,4,8,16,32.

Question:4 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

     a _n = \frac{2n-3 }{6}

Answer:

Given : a _n = \frac{2n-3 }{6}

a _1 = \frac{2\times 1-3 }{6}=\frac{-1}{6}

a _2 = \frac{2\times 2-3 }{6}=\frac{1}{6}

a _3 = \frac{2\times 3-3 }{6}=\frac{3}{6}=\frac{1}{2}

a _4 = \frac{2\times 4-3 }{6}=\frac{5}{6}

a _5 = \frac{2\times 5-3 }{6}=\frac{7}{6}

Therefore, the required number of terms =\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}

Question:5 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: 

     a _ n = ( -1) ^{n-1} 5 ^{n+1}

Answer:

Given : a _ n = ( -1) ^{n-1} 5 ^{n+1}

a _ 1 = ( -1) ^{1-1} 5 ^{1+1}=(-1)^{0}.5^2=25

a _ 2 = ( -1) ^{2-1} 5 ^{2+1}=(-1)^{1}.5^3=-125

a _ 3 = ( -1) ^{3-1} 5 ^{3+1}=(-1)^{2}.5^4= 625

a _ 4 = ( -1) ^{4-1} 5 ^{4+1}=(-1)^{3}.5^5= -3125

a _ 5 = ( -1) ^{5-1} 5 ^{5+1}=(-1)^{4}.5^6= 15625

Therefore, the required number of terms =25,-125,625,-3125,15625

Question:6 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

    a _n = n \frac{n^2 + 5}{4}

Answer:

Given : a _n = n \frac{n^2 + 5}{4}

a _1 = 1. \frac{1^2 + 5}{4}=\frac{6}{4}=\frac{3}{2}

a _2 = 2. \frac{2^2 + 5}{4}=\frac{18}{4}=\frac{9}{2}

a _3 = 3. \frac{3^2 + 5}{4}=\frac{42}{4}=\frac{21}{2}

a _4 = 4. \frac{4^2 + 5}{4}=\frac{84}{4}=21

a _5 = 5. \frac{5^2 + 5}{4}=\frac{150}{4}=\frac{75}{2}

Therefore, the required number of terms =\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}

Question:12 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

       a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2

Answer:

Given : a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2

a _ 2 = \frac{a_{2-1}}{2} =\frac{a_1}{2}=\frac{-1}{2}

a _ 3 = \frac{a_{3-1}}{3} =\frac{a_2}{3}=\frac{-1}{6}

a _ 4 = \frac{a_{4-1}}{4} =\frac{a_3}{4}=\frac{-1}{24}

a _ 5 = \frac{a_{5-1}}{5} =\frac{a_4}{5}=\frac{-1}{120}

Hence, five terms of series are -1,\frac{-1}{2},\frac{-1}{-6},\frac{-1}{24},\frac {-1}{120}

Series  

               =-1+\frac{-1}{2}+\frac{-1}{-6}+\frac{-1}{24}+\frac {-1}{120}.........................

Question:13 Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: a_1 = a_2 = 2, a_n = a_{n - 1}-1, n > 2

Answer:

Given : a_1 = a_2 = 2, a_n = a_{n - 1}-1, n > 2

a_3 = a_{3 - 1}-1=a_2-1=2-1=1

a_4 = a_{4 - 1}-1=a_3-1=1-1=0

a_5 = a_{5 - 1}-1=a_4-1=0-1=-1

Hence, five terms of series are 2,2,1,0,-1

Series =2+2+1+0+(-1)+..................

Question:14 The Fibonacci sequence is defined by 1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2

        Find\frac{a _{n+1}}{a_n}, for n = 1, 2, 3, 4, 5

Answer:

Given : The Fibonacci sequence is defined by 1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2

a _3 = a _{3-1} + a _{3-2} =a_2+a_1=1+1=2

a _4 = a _{4-1} + a _{4-2} =a_3+a_2=2+1=3

a _5 = a _{5-1} + a _{5-2} =a_4+a_3=3+2=5

a _6 = a _{6-1} + a _{6-2} =a_5+a_4=5+3=8

 For \,\,n=1,\frac{a _{n+1}}{a_n}=\frac {a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1

For \,\, n=2,\frac{a _{n+1}}{a_n}=\frac {a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2

For \,\, n=3,\frac{a _{n+1}}{a_n}=\frac {a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}

For \,\, n=4,\frac{a _{n+1}}{a_n}=\frac {a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}

For \,\, n=5,\frac{a _{n+1}}{a_n}=\frac {a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}

 

Solutions of NCERT for class 11 maths chapter 9 sequences and series-Exercise: 9.2

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

Answer:

Odd integers from 1 to 2001 are 1,3,5,7...........2001.

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know , a_n = a+(n-1)d

              2001 = 1+(n-1)2

      \Rightarrow \, \, 2000 = (n-1)2

    \Rightarrow \, \, 1000 = (n-1)

 \Rightarrow \, \, n=1000+1=1001

 

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

       = \frac{1001}{2}[2(1)+(1001-1)2]

     = \frac{1001}{2}[2002]

    = 1001\times 1001

   = 1002001

The , sum of odd integers from 1 to 2001 is 1002001.

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

Answer:

Numbers divisible by 5 from 100 to 1000 are 105,110,.............995

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know , a_n = a+(n-1)d

              995 = 105+(n-1)5

      \Rightarrow \, \, 890 = (n-1)5

    \Rightarrow \, \, 178 = (n-1)

 \Rightarrow \, \, n=178+1=179

 

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

       = \frac{179}{2}[2(105)+(179-1)5]

     = \frac{179}{2}[2(105)+178(5)]

      = 179\times 550

     = 98450

The  sum of numbers divisible by 5 from 100 to 1000 is 98450.

Question:3 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. 

Answer:

First term =a=2

Let the series be 2,2+d,2+2d,2+3d,.......................

Sum of first five terms =10+10d

Sum of next five terms =10+35d

Given : The sum of the first five terms is one-fourth of  the next five terms.

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

\Rightarrow \, \, 40+40d=10+35d

\Rightarrow \, \, 40-10=35d-40d

\Rightarrow \, \, 30=-5d

\Rightarrow \, \, d=-6

To prove  : a_2_0=-112

L.H.S : a_2_0=a+(20-1)d=2+(19)(-6)=2-114=-112=R.H.S

Hence, 20th term is –112.

Question:4 How many terms of the A.P. -6 , -11/2 , -5...  are needed to give the sum –25?

Answer:

Given : A.P. = -6 , -11/2 , -5...

a=-6

d=\frac{-11}{2}+6=\frac{1}{2}

Given : sum = -25

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

\Rightarrow \, \, -25=\frac{n}{2}[2(-6)+(n-1)\frac{1}{2}]

\Rightarrow \, \, \, \, -50= n[-12+(n-1)\frac{1}{2}]

\Rightarrow \, \, \, \, -50= -12n+ \frac{n^2}{2}-\frac{n}{2}

\Rightarrow \, \, \, \, -100= -24n+ n^2-n

\Rightarrow \, \, \, \, n^2-25n+100=0

\Rightarrow \, \, \, \, n^2-5n-20n+100=0

\Rightarrow \, \, \, \, n(n-5)-20(n-5)=0

\Rightarrow \, \, \, \, (n-5)(n-20)=0

\Rightarrow \, \, \, \, n=5\, \, or\, \, 20.

Question:5 In an A.P., if pth term is 1/q  and qth term is 1/p  , prove that the sum of first pq terms is 1/2  (pq +1), where p \neq q

Answer:

Given : In an A.P., if pth term is 1/q  and qth term is 1/p 

a_p=a+(p-1)d=\frac{1}{q}.................(1)

a_q=a+(q-1)d=\frac{1}{p}.................(2)

Subtracting (2) from (1), we get

           \Rightarrow \, \, a_p-a_q

 \Rightarrow \, \, (p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}

\Rightarrow \, \, pd-d-qd+d=\frac{p-q}{pq}

\Rightarrow \, \, (p-q)d=\frac{p-q}{pq}

\Rightarrow \, \, d=\frac{1}{pq}

 Putting value of d in equation (1),we get 

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

\Rightarrow a+\frac{1}{q}-\frac{1}{pq} = \frac{1}{q}

\Rightarrow a= \frac{1}{pq}

\therefore \, \, S_p_q=\frac{pq}{2}[2.\frac{1}{pq}+(pq-1).\frac{1}{pq}]

\Rightarrow \, \, S_p_q=\frac{1}{2}[2+(pq-1)]

\Rightarrow \, \, S_p_q=\frac{1}{2}[pq+1]

Hence,the sum of first pq  terms is 1/2  (pq +1), where p \neq q.

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

Answer:

Given : A.P. 25, 22, 19, ….....

S_n=116

a=25  , d = -3

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

\Rightarrow \, \, 116=\frac{n}{2}[2(25)+(n-1)(-3)]

\Rightarrow \, \, 232=n[50-3n+3]

\Rightarrow \, \, 232=n[53-3n]

\Rightarrow \, \, 3n^2-53n+232=0

\Rightarrow \, \, 3n^2-24n-29n+232=0

\Rightarrow \, \, 3n(n-8)-29(n-8)=0

\Rightarrow \, \, (3n-29)(n-8)=0

\Rightarrow \, \, n=8\, \, or\, \, \, n=\frac{29}{3}

n could not be \frac{29}{3} so n=8.

Last term =a_8=a+(n-1)d

                        =25+(8-1)(-3)

                      =25-21=4

The, last term of A.P. is 4.

Question:7 Find the sum to n terms of the A.P., whose k^{th} term is 5k + 1.

Answer:

Given : a_k=5k+1

\Rightarrow \, \, a+(k-1)d=5k+1

\Rightarrow \, \, a+kd-d=5k+1

Comparing LHS and RHS , we have

a-d=1        and     d=5

Putting value of d,

a=1+5=6

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

S_n=\frac{n}{2}[2(6)+(n-1)5]

S_n=\frac{n}{2}[12+5n-5]

S_n=\frac{n}{2}[7+5n]

Question:8 If the sum of n terms of an A.P. is ( pn + qn ^ 2 ) , where p and q are constants,  find the common difference

Answer:

If the sum of n terms of an A.P. is ( pn + qn ^ 2 ),

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

\Rightarrow \, \, \frac{n}{2}[2a+(n-1)d]=pn+qn^2

\Rightarrow \, \, \frac{n}{2}[2a+nd-d]=pn+qn^2

\Rightarrow \, \, an+\frac{n^2}{2}d-\frac{nd}{2}=pn+qn^2

Comparing coefficients of n^2 on both side , we get

          \frac{d}{2}=q

     \Rightarrow \, \, d=2q

The common difference of AP is 2q.

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

Answer:

Given: The sums of n terms of two arithmetic progressions are in the ratio.5n + 4 : 9n + 6

There are two AP's with first terms =a_1,a_2    and common difference =  d_1,d_2

\Rightarrow \, \, \frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{5n+4}{9n+6}

\Rightarrow \, \, \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}

Substituting n=35,we get

\Rightarrow \, \, \frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_2}=\frac{5(35)+4}{9(35)+6}

\Rightarrow \, \, \frac{2a_1+34 d_1}{2a_2+34d_2}=\frac{5(35)+4}{9(35)+6}

\Rightarrow \, \, \frac{a_1+17 d_1}{a_2+17d_2}=\frac{179}{321}

\Rightarrow \, \, \frac{18^t^h \, term \, of\, first \, AP}{18^t^h\, term\, of\, second\, AP}=\frac{179}{321}

Thus, the ratio of the 18th term of AP's is 179:321

Question:10 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.

Answer:

Let first term of AP = a and common difference = d.

Then,

S_p=\frac{p}{2}[2a+(p-1)d]

S_q=\frac{q}{2}[2a+(q-1)d]

 Given : S_p=S_q

\Rightarrow \frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]

\Rightarrow p[2a+(p-1)d]=q[2a+(q-1)d]

\Rightarrow 2ap+p^2d-pd=2aq+q^2d-qd

\Rightarrow 2ap+p^2d-pd-2aq-q^2d+qd=0

\Rightarrow 2a(p-q)+d(p^2-p-q^2+q)=0

\Rightarrow 2a(p-q)+d((p-q)(p+q)-(p-q))=0

\Rightarrow 2a(p-q)+d[(p-q)(p+q-1)]=0

\Rightarrow (p-q)[2a+d(p+q-1)]=0

\Rightarrow 2a+d(p+q-1)=0

\Rightarrow d(p+q-1)=-2a

\Rightarrow d=\frac{-2a}{p+q-1}

 

Now, S_(_p_+_q_)= \frac{p+q}{2}[2a+(p+q-1)d]

                    =\frac{p+q}{2}[2a+(p+q-1)\frac{-2a}{p+q-1}]

                  =\frac{p+q}{2}[2a+(-2a)]

                 =\frac{p+q}{2}[0]=0

Thus, sum of p+q terms of AP is 0.

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

      \frac{a}{p} ( q-r ) + \frac{b}{q}( r-p ) + \frac{c}{r} ( p-q ) = 0

Answer:

To prove : \frac{a}{p} ( q-r ) + \frac{b}{q}( r-p ) + \frac{c}{r} ( p-q ) = 0

Let a_1 and d be the first term and the common difference of AP, respectively.

According to the given information, we have

S_p=\frac{p}{2}[2a_1+(p-1)d]=a

\Rightarrow [2a_1+(p-1)d]=\frac{2a}{p}............(1)

 

S_q=\frac{q}{2}[2a_1+(q-1)d]=b

\Rightarrow [2a_1+(q-1)d]=\frac{2b}{q}............(2)

 

S_r=\frac{r}{2}[2a_1+(r-1)d]=c

\Rightarrow [2a_1+(r-1)d]=\frac{2c}{r}............(3)

 

Subtracting equation (2) from (1), we have 

\Rightarrow (p-1)d-(q-1)d=\frac{2a}{p}-\frac{2b}{q}

\Rightarrow d(p-q-1+1)=\frac{2(aq-bp)}{pq}

\Rightarrow d(p-q)=\frac{2(aq-bp)}{pq}

\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}

Subtracting equation (3) from (2), we have 

\Rightarrow (q-1)d-(r-1)d=\frac{2b}{q}-\frac{2c}{r}

\Rightarrow d(q-r-1+1)=\frac{2(br-cq)}{qr}

\Rightarrow d(q-r)=\frac{2(br-qc)}{qr}

\Rightarrow d=\frac{2(br-qc)}{qr(q-r)}

Equating values of d, we have 

\Rightarrow d=\frac{2(aq-bp)}{pq(p-q)}=\frac{2(br-qc)}{qr(q-r)}

\Rightarrow \frac{2(aq-bp)}{pq(p-q)}=\frac{2(br-qc)}{qr(q-r)}

\Rightarrow \, \, (aq-bp)qr(q-r)=(br-qc)pq(p-q)

\Rightarrow \, \, (aq-bp)r(q-r)=(br-qc)p(p-q)

\Rightarrow \, \, (aqr-bpr)(q-r)=(bpr-pqc)(p-q)

Dividing both sides from pqr, we get 

\Rightarrow \, \, (\frac{a}{p}-\frac{b}{q})(q-r)=(\frac{b}{q}-\frac{c}{r})(p-q)

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

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

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

Hence, the given result is proved.

Question:12 The ratio of the sums of m and n terms of an A.P. is m^2 : n^2 . Show that the ratio of mth and nth term is ( 2m-1) : ( 2n- 1 ).

Answer:

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is m^2 : n^2 .

To prove :  the ratio of mth and nth term is ( 2m-1) : ( 2n- 1 ).

 \therefore \, \frac{sum\, of\, m\, \, terms}{sum\, of\, n\, \, terms }=\frac{m^2}{n^2}

\Rightarrow \, \, \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}

\Rightarrow \, \, \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}

Put m=2m-1\, \, and\, \, n=2n-1, we get 

\Rightarrow \, \, \frac{2a+(2m-2)d}{2a+(2n-2)d}=\frac{2m-1}{2n-1}

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

\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{a+(m-1)d}{a+(n-1)d}

From equation (1) ,we get

\Rightarrow \, \, \frac{m\, th \, \, term\, \, of\, AP}{n\, th\, \, term\, \, of\, \, AP}=\frac{2m-1}{2n-1}

Hence proved.

Question:13 If the sum of n terms of an A.P. is  3 n^2 + 5 n  and its m^{th } term is 164, find the value of m.

Answer:

Given : If the sum of n terms of an A.P. is  3 n^2 + 5 n  and its m^{th } term is 164

Let a and d be first term and common difference of a AP ,respectively.

 Sum of n terms = 3 n^2 + 5 n

\Rightarrow \, \, \frac{n}{2}[2a+(n-1)d]=3n^2+5n

\Rightarrow \, \, 2a+(n-1)d=6n+10

\Rightarrow \, \, 2a+nd-d=6n+10

Comparing the coefficients of n on both side , we have

     \Rightarrow \, \, d=6

Also , 2a-d=10

\Rightarrow \, \, 2a-6=10

\Rightarrow \, \, 2a=10+6

\Rightarrow \, \, 2a=16

\Rightarrow \, \, a=8

 m th term is 164.

\Rightarrow \, \, a+(m-1)d=164

\Rightarrow \, \, 8+(m-1)6=164

\Rightarrow \, \, (m-1)6=156

\Rightarrow \, \, m-1=26

\Rightarrow \, \, m=26+1=27

Hence, the value of m is 27.

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

Answer:

Let five numbers be A,B,C,D,E.

Then  AP=8,A,B,C,D,E,26

Here we have,

      a=8,a_7=26,n=7

 \Rightarrow \, \, a+(n-1)d=a_n

\Rightarrow \, \, 8+(7-1)d=26

\Rightarrow \, \, 6d=18

\Rightarrow \, \, d=\frac{18}{6}=3

Thus, we have A=a+d=8+3=11

                      B=a+2d=8+(2)3=8+6=14

                     C=a+3d=8+(3)3=8+9=17

                   D=a+4d=8+(4)3=8+12=20

                  E=a+5d=8+(5)3=8+15=23

Thus, the five numbers are 11,14,17,20,23.

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

Answer:

Given :  \frac{a^n + b ^n }{a ^{ n-1}+ b ^{n-1}}  is the A.M. between a and b.

\frac{a^n + b ^n }{a ^{ n-1}+ b ^{n-1}}=\frac{a+b}{2}

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

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

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

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

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

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

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

\Rightarrow \,\left [ \frac{a}{b} \right ]^{n-1}= 1

\Rightarrow \,n-1=0

\Rightarrow \,n=1

Thus, value of n is 1.

Question: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 7 ^{th} and  (m-1)^{th}  numbers is 5 : 9. Find the  value of m.

Answer:

Let A,B,C.........M be m numbers.

Then, AP=1,A,B,C..........M,31

Here we have,

  a=1,a_m_+_2=31,n=m+2

     \Rightarrow \, \, a+(n-1)d=a_n

\Rightarrow \, \, 1+(m+2-1)d=31

\Rightarrow \, \, (m+1)d=30

\Rightarrow \, \, d=\frac{30}{m+1}

Given : the ratio of 7 ^{th} and  (m-1)^{th}  numbers is 5 : 9.

\Rightarrow \, \, \frac{a+(7)d}{a+(m-1)d}=\frac{5}{9}

\Rightarrow \, \, \frac{1+7d}{1+(m-1)d}=\frac{5}{9}

\Rightarrow \, \, 9(1+7d)=5(1+(m-1)d)

\Rightarrow \, \, 9+63d=5+5md-5d

Putting value of d from above,

\Rightarrow \, \, 9+63(\frac{30}{m+1})=5+5m\left ( \frac{30}{m+1} \right )-5\left ( \frac{30}{m+1} \right )

\Rightarrow \, \9(m+1)+1890=5(m+1)+150m-150

\Rightarrow \, \9m+9+1890=5m+5+150m-150

\Rightarrow \, 1890+9-5+150=155m-9m

\Rightarrow \, 2044=146m

\Rightarrow \, m=14

Thus, value of m is 14.

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

Answer:

The  first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

AP= 100,105,110,115,.................

We have ,a=100\, \, and\, \, \, d=5

a_n=a+(n-1)d

a_3_0=100+(30-1)5

a_3_0=100+(29)5

a_3_0=100+145

a_3_0=245

Thus, he will pay  Rs. 245 in the 30th instalment.

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

Answer:

The angles of polygon forms AP with common difference of 5 \degree and first term as 120 \degree .

We know that sum of angles of polygon with n sides is 180(n-2)

\therefore S_n=180(n-2)

\Rightarrow \frac{n}{2}[2a+(n-1)d]=180(n-2)

\Rightarrow \frac{n}{2}[2(120)+(n-1)5]=180(n-2)

\Rightarrow n[240+5n-5]=360n-720

\Rightarrow 235n+5n^2=360n-720

\Rightarrow 5n^2-125n+720=0

\Rightarrow n^2-25n+144=0

\Rightarrow n^2-16n-9n+144=0

\Rightarrow n(n-16)-9(n-16)=0

\Rightarrow (n-16)(n-9)=0

\Rightarrow n=9,16

Sides of polygon are 9 or 16.

 

CBSE NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.3

Question:1 Find the 20 ^{th} and n ^{th} terms of the G.P. \frac{5}{2},\frac{5}{4},\frac{5}{8},....

Answer:

G.P : 

            \frac{5}{2},\frac{5}{4},\frac{5}{8},....

first term = a

   a=\frac{5}{2}

common ratio =r

r=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}

a_n=a.r^{n-1}

a_2_0=\frac{5}{2}.(\frac{1}{2})^{20-1}

a_2_0=\frac{5}{2}.(\frac{1}{2^{19}})

a_2_0=\frac{5}{2^{20}}

a_n=a.r^{n-1}

a_n=\frac{5}{2}.\left ( \frac{1}{2} \right )^{n-1}

a_n=\frac{5}{2}. \frac{1}{2^{n-1}}

a_n=\frac{5}{2^n}   the nth term of G.P

Question:2 Find the 12 ^{th} term of a G.P. whose 8 ^{th} term is 192 and the common ratio is 2.

Answer:

First term = a

common ratio =r=2

8 ^{th} term is 192

a_n=a.r^{n-1}

a_8=a.(2)^{8-1}

192=a.(2)^{7}

a=\frac{2^6.3}{2^7}

a=\frac{3}{2}

a_n=a.r^{n-1}

a_1_2=\frac{3}{2}. ( 2 )^{12-1}

a_1_2=\frac{3}{2}. ( 2 )^{11}

a_1_2= 3. ( 2 )^{10}

a_1_2= 3072 is the 12 ^{th} term of a G.P.

Question:3 The 5 ^{th} , 8 ^{th} \: \:and \: \: 11 ^{th}  terms of a G.P. are p, q and s, respectively. Show  that q ^2 = ps

Answer:

To prove : q ^2 = ps

Let first term=a and common ratio = r

a_5=a.r^4=p..................(1)

a_8=a.r^7=q..................(2)

a_1_1=a.r^1^0=s..................(3)

Dividing equation 2 by 1, we have

\frac{a.r^7}{a.r^4}=\frac{q}{p}

\Rightarrow r^3=\frac{q}{p}

Dividing equation 3 by 2, we have

\frac{a.r^1^0}{a.r^7}=\frac{s}{q}

\Rightarrow r^3=\frac{s}{q}

Equating values of r^3 ,  we have

\frac{q}{p}=\frac{s}{q}

\Rightarrow q^2=ps

Hence proved

Question:4 The 4^{th} term of a G.P. is square of its second term, and the first term is -3. Determine its 7^{th} term.

Answer:

First term =a= -3

4^{th} term of a G.P. is square of its second term

\Rightarrow a_4=(a_2)^2

\Rightarrow a.r^{4-1}=(a.r^{2-1})^2

\Rightarrow a.r^{3}=a^2.r^{2}

\Rightarrow r=a=-3

a_7=a.r^{7-1}

\Rightarrow a_7=(-3).(-3)^{6}

\Rightarrow a_7=(-3)^{7}=-2187

Thus, seventh term is -2187.

Question:5(a) Which term of the following sequences:  2,2\sqrt 2 , 4 .,....is \: \: 128 ?

Answer:

Given :  GP = 2,2\sqrt 2 , 4 .,............

a=2\, \, \, \, \, and \, \, \, \, \, r=\frac{2\sqrt{2}}{2}=\sqrt{2}

n th term is given as 128.

a_n=a.r^{n-1}

\Rightarrow 128=2.(\sqrt{2})^{n-1}

\Rightarrow 64=(\sqrt{2})^{n-1}

\Rightarrow 2^6=(\sqrt{2})^{n-1}

\Rightarrow \sqrt{2}^1^2=(\sqrt{2})^{n-1}

\Rightarrow n-1=12

\Rightarrow n=12+1=13

The, 13 th term is 128.

Question:5(b) Which term of the following sequences: \sqrt 3 ,3 , 3 \sqrt 3 ,...is \: \: 729 ?

Answer:

Given :  GP=\sqrt 3 ,3 , 3 \sqrt 3 ,........

a=\sqrt{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{3}{\sqrt{3}}=\sqrt{3}

n th term is given as 729.

a_n=a.r^{n-1}

\Rightarrow 729=\sqrt{3}.(\sqrt{3})^{n-1}

\Rightarrow 729=(\sqrt{3})^{n}

\Rightarrow( \sqrt{3})^1^2=(\sqrt{3})^{n}

\Rightarrow n=12

The, 12 th term is 729.

Question:5(c) Which term of the following sequences:  \frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,....is \: \: \frac{1}{19683}?

Answer:

Given :  GP=\frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,............

a=\frac{1}{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}

n th term is given as  \frac{1}{19683}

a_n=a.r^{n-1}

\Rightarrow \frac{1}{19683}=\frac{1}{3}.(\frac{1}{3})^{n-1}

\Rightarrow \frac{1}{19683}=\frac{1}{3^n}

\Rightarrow \frac{1}{3^9}=\frac{1}{3^n}

\Rightarrow n=9

Thus, n=9.

Question:6 For what values of x, the numbers -\frac{2}{7} ,x, -\frac{7}{2}  are in G.P.?

Answer:

GP=-\frac{2}{7} ,x, -\frac{7}{2}

Common ratio=r.

r=\frac{x}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{x}

\Rightarrow x^2=1

\Rightarrow x=\pm 1

Thus, for x=\pm 1 ,given numbers will be in GP.

Question:7 Find the sum to indicated number of terms in each of the geometric progressions in  0.15, 0.015, 0.0015, ... 20 terms.

Answer:

geometric progressions is  0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

S_n=\frac{a(1-r^n)}{1-r}

S_2_0=\frac{0.15(1-(0.1)^{20})}{1-0.1}

S_2_0=\frac{0.15(1-(0.1)^{20})}{0.9}

S_2_0=\frac{0.15}{0.9}(1-(0.1)^{20})

S_2_0=\frac{15}{90}(1-(0.1)^{20})

S_2_0=\frac{1}{6}(1-0.1^{20})

Question:9 Find the sum to indicated number of terms in each of the geometric progressions in -a , a^2 , - a ^3 , ... n terms ( if a \neq -1)

Answer:

The sum to the indicated number of terms in each of the geometric progressions is:

GP=1,-a , a^2 , - a ^3 , .............

a=1\, \, \, and\, \, \, \, r=-a

S_n=\frac{a(1-r^n)}{1-r}

S_n=\frac{1(1-(-a)^n)}{1-(-a)}

S_n=\frac{1(1-(-a)^n)}{1+a}

S_n=\frac{1-(-a)^n}{1+a}

Question:11 Evaluate     \sum_{k = 1}^{11} ( 2+ 3 ^k ) 

Answer:

Given :   

              \sum_{k = 1}^{11} ( 2+ 3 ^k )

\sum_{k = 1}^{11} ( 2+ 3 ^k )=\sum _{k=1}^{11}2 +\sum _{k=1}^{11} 3^k

                          =22 +\sum _{k=1}^{11} 3^k...............(1)

\sum _{k=1}^{11} 3^k=3^1+3^2+3^3+....................3^1^1

These terms form GP with a=3 and r=3.

S_n=\frac{a(1-r^n)}{1-r}

S_n=\frac{3(1-3^1^1)}{1-3}

S_n=\frac{3(1-3^1^1)}{-2}

S_n=\frac{3(3^1^1-1)}{2}=\sum _{k=1}^{11} 3^k

\sum_{k = 1}^{11} ( 2+ 3 ^k )=22+\frac{3(3^1^1-1)}{2}

Question:12 The sum of first three terms of a G.P. is \frac{39}{10} and their product is 1. Find the common ratio and the terms.

Answer:

Given : The sum of first three terms of a G.P. is \frac{39}{10} and their product is 1.

Let three terms be   \frac{a}{r},a,ar.

S_n=\frac{a(1-r^n)}{1-r}

S_3=\frac{a(1-r^3)}{1-r}=\frac{39}{10}

\frac{a}{r}+a+ar=\frac{39}{10}.........1

Product of 3 terms is 1.

\frac{a}{r}\times a\times ar=1

\Rightarrow a^3=1

\Rightarrow a=1

Put value of a in equation 1, 

\frac{1}{r}+1+r=\frac{39}{10}

10(1+r+r^2)=39(r)

\Rightarrow 10r^2-29r+10=0

\Rightarrow 10r^2-25r-4r+10=0

\Rightarrow 5r(2r-5)-2(2r-5)=0

\Rightarrow (2r-5)(5r-2)=0

\Rightarrow r=\frac{5}{2},r=\frac{2}{5}

The three terms of AP are \frac{5}{2},1,\frac{2}{5}.

Question:13 How many terms of G.P. 3 , 3 ^ 2 , 3 ^ 3, … are needed to give the sum 120? 

Answer:

G.P.=    3 , 3 ^ 2 , 3 ^ 3, …............

Sum =120

These terms are GP with a=3 and r=3.

S_n=\frac{a(1-r^n)}{1-r}

120=\frac{3(1-3^n)}{1-3}

120\times \frac{-2}{3}=(1-3^n)

-80=(1-3^n)

\Rightarrow 3^n=1+80=81

\Rightarrow 3^n=81

\Rightarrow 3^n=3^4

\Rightarrow n=4

Hence, we have value of n as 4 to get sum of 120.

Question:14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer:

Let GP be a,ar,ar^2,ar^3,ar^4,ar^5,ar^6................................

Given :  The sum of first three terms of a G.P. is 16

           a+ar+ar^2=16

\Rightarrow a(1+r+r^2)=16...............................(1)

Given :  the sum of the next three terms is128.

       ar^3+ar^4+ar^5=128

\Rightarrow ar^3(1+r+r^2)=128...............................(2)

Dividing equation (2) by (1), we have

\Rightarrow \frac{ar^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}

\Rightarrow r^3=8

\Rightarrow r^3=2^3

\Rightarrow r=2

Putting value of r =2 in equation 1,we have 

\Rightarrow a(1+2+2^2)=16

\Rightarrow a(7)=16

\Rightarrow a=\frac{16}{7}

S_n=\frac{a(1-r^n)}{1-r}

S_n=\frac{\frac{16}{7}(1-2^n)}{1-2}

S_n=\frac{16}{7}(2^n-1)

Question:15 Given a G.P. with a = 729 and 7 ^{th} term 64, determine s_7

Answer:

Given a G.P. with a = 729 and 7 ^{th} term 64.

a_n=a.r^{n-1}

\Rightarrow 64=729.r^{7-1}

\Rightarrow r^6=\frac{64}{729}

\Rightarrow r^6=\left ( \frac{2}{3} \right )^6

\Rightarrow r=\frac{2}{3}

S_n=\frac{a(1-r^n)}{1-r}

S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{1-\frac{2}{3}}

S_7=\frac{729(1-\left ( \frac{2}{3} \right )^7)}{\frac{1}{3}}

S_7=3\times 729 \left ( \frac{3^7-2^7}{3^7} \right )

S_7= \left ( 3^7-2^7 \right )

S_7= 2187-128

S_7= 2059 (Answer)

Question:16 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term

Answer:

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

a_5=4.a_3

\Rightarrow a.r^{5-1}=4.a.r^{3-1}

\Rightarrow a.r^{4}=4.a.r^{2}

\Rightarrow r^{2}=4

\Rightarrow r=\pm 2

If r=2, then

S_n=\frac{a(1-r^n)}{1-r}

\Rightarrow \frac{a(1-2^2)}{1-2}=-4

\Rightarrow \frac{a(1-4)}{-1}=-4

\Rightarrow a(-3)=4

\Rightarrow a=\frac{-4}{3}

 

If r= - 2, then

S_n=\frac{a(1-r^n)}{1-r}

\Rightarrow \frac{a(1-(-2)^2)}{1-(-2)}=-4

\Rightarrow \frac{a(1-4)}{3}=-4

\Rightarrow a(-3)=-12

\Rightarrow a=\frac{-12}{-3}=4

Thus, required GP is \frac{-4}{3},\frac{-8}{3},\frac{-16}{3},.........      or    4,-8,-16,-32,..........

Question:17 If the 4 ^{th} , 10 ^{th} , 16 ^ {th}  terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.

Answer:

Let x,y, z are in G.P.

Let first term=a and common ratio = r

a_4=a.r^3=x..................(1)

a_1_0=a.r^9=y..................(2)

a_1_6=a.r^1^5=z..................(3)

Dividing equation 2 by 1, we have

\frac{a.r^9}{a.r^3}=\frac{y}{x}

\Rightarrow r^4=\frac{y}{x}

Dividing equation 3 by 2, we have

\frac{a.r^1^5}{a.r^9}=\frac{z}{y}

\Rightarrow r^4=\frac{z}{y}

Equating values of r^4 ,  we have

\frac{y}{x}=\frac{z}{y}

Thus, x,y,z are in GP

Question:18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Answer:

 8, 88, 888, 8888… is not a GP.

It can be changed in GP by writing terms as 

S_n=8+88+888+8888+............. to n terms

S_n=\frac{8}{9}[9+99+999+9999+................]

S_n=\frac{8}{9}[(10-1)+(10^2-1)+(10^3-1)+(10^4-1)+................]

S_n=\frac{8}{9}[(10+10^2+10^3+........)-(1+1+1.....................)]

S_n=\frac{8}{9}[\frac{10(10^n-1)}{10-1}-(n)]

S_n=\frac{8}{9}[\frac{10(10^n-1)}{9}-(n)]

S_n=\frac{80}{81}(10^n-1)-\frac{8n}{9}

Question:19 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,1/2

Answer:

Required \, \, sum=2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}

Required \, \, sum=64\left [ 4+2+1+\frac{1}{2}+\frac{1}{2^2} \right ]

    Here, 4,2,1,\frac{1}{2},\frac{1}{2^2}    is a GP.

first term =a=4

common ratio =r 

            r=\frac{1}{2}      

S_n=\frac{a(1-r^n)}{1-r}

S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{1-\frac{1}{2}}

S_5=\frac{4(1-\left ( \frac{1}{2} \right )^5)}{\frac{1}{2}}

S_5=8(1-\left ( \frac{1}{32} \right ))

S_5=8(\frac{31}{32})

S_5=\frac{31}{4}

Required \, \, sum=64\left [ \frac{31}{4} \right ]

Required \, \, sum=16\times 31=496

Question:20 Show that the products of the corresponding terms of the sequences a,ar, ar^2 , ...ar^{n-1} \: \: and\: \: A ,AR, AR^2 ....AR^{n-1} form a G.P, and find the common ratio.

Answer:

To prove : aA,arAR,ar^2AR^2,...................     is a GP.

\frac{second \, \, term}{first\, \, term}=\frac{arAR}{aA}=rR

\frac{third \, \, term}{second\, \, term}=\frac{ar^2AR^2}{arAR}=rR

Thus, the above sequence is a GP with common ratio of rR.

Question:21  Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and  the second term is greater than the 4 ^{th} by 18.

Answer:

Let first term be a and common ratio be r.

a_1=a,a_2=ar,a_3=ar^2,a_4=ar^3

Given : the third term is greater than the first term by 9, and  the second term is greater than the 4 ^{th} by 18.

a_3=a_1+9

\Rightarrow ar^2=a+9

\Rightarrow a(r^2-1)=9.................1

a_2=a_4+18

\Rightarrow ar=ar^3+18

\Rightarrow ar(1-r^2)=18......................2

Dividing equation 2 by 1 , we get

\frac{ ar(1-r^2)}{ -a(1-r^2)}=\frac{18}{9}

\Rightarrow r=-2

Putting value of r , we get

4a=a+9

\Rightarrow 4a-a=9

\Rightarrow 3a=9

\Rightarrow a=3

Thus, four terms of GP are 3,-6,12,-24.

Question:22 If the p^{th} , q ^{th} , r ^{th} terms of a G.P. are a, b and c, respectively. Prove that  a ^{ q-r } b ^{r- p } C ^{p-q} = 1

Answer:

To prove : a ^{ q-r } b ^{r- p } C ^{p-q} = 1

Let A  be the first term and R be common ratio.

According to the given information, we have

a_p=A.R^{p-1}=a

a_q=A.R^{q-1}=b

a_r=A.R^{r-1}=c

L.H.S : a ^{ q-r } b ^{r- p } C ^{p-q}

           =A^{q-r}.R^{(q-r)(p-1)}.A^{r-p}.R^{(r-p)(q-1)}.A^{p-q}.R^{(p-q)(r-1)}

          =A^{q-r+r-p+p-q}.R^{(qp-rp-q+r)+(rq-pq+p-r)+(pr-p-qr+q)}

         =A^0.R^0=1=RHS

Thus, LHS = RHS.

Hence proved.

Question:23 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that  P^2 = ( ab)^n.

Answer:

Given : First term =a  and n th term = b.

Common ratio = r.

To prove : P^2 = ( ab)^n

Then , GP = a,ar,ar^2,ar^3,ar^4,..........................

a_n=a.r^{n-1}=b..................................1

P = product of n terms

P=(a).(ar).(ar^2).(ar^3)..............(ar^{n-1})

P=(a.a.a...............a)((1).(r).(r^2).(r^3)..............(r^{n-1}))

P=(a^n)(r^{1+2+.........(n-1)})........................................2

Here, 1+2+.........(n-1)   is a AP.

\therefore\, \, \, sum= \frac{n}{2}\left [2a+(n-1)d \right ]

                   = \frac{n-1}{2}\left [2(1)+(n-1-1)1 \right ]

                   = \frac{n-1}{2}\left [2+n-2 \right ]

                    = \frac{n-1}{2}\left [n \right ]

                   = \frac{n(n-1)}{2}

Put in equation (2),

P=(a^n)(r^{\frac{n(n-1)}{2}})

P^2=(a^2^n)(r^{n(n-1)})

P^2=(a. a.r^{(n-1)})^n

P^2=(a.b)^n

 Hence proved .

Question:24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from  ( n+1)^{th} \: \: to\: \: (2n)^{th} term is \frac{1}{r^n}

Answer:

Let first term =a  and common ratio = r.

sum \, \, of\, \, n\, \, terms=\frac{a(1-r^n)}{1-r}

Since there are n terms from (n+1) to 2n  term.

Sum of terms from (n+1) to 2n.

S_n=\frac{a_(_n+_1_)(1-r^n)}{1-r}

a_(_n+_1)=a.r^{n+1-1}=ar^n

Thus, the required ratio  = \frac{a(1-r^n)}{1-r}\times \frac{1-r}{ar^n(1-r^n)}

                                =\frac{1}{r^n}

Thus,  the common ratio of the sum of first n terms of a G.P. to the sum of terms from  ( n+1)^{th} \: \: to\: \: (2n)^{th} term is  \frac{1}{r^n}.

Question:25 If a, b, c and d are in G.P. show that    (a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .

Answer:

 If a, b, c and d are in G.P.

bc=ad....................(1)

b^2=ac....................(2)

c^2=bd....................(3)

To prove : (a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .

RHS : (ab + bc + cd)^2 .

       =(ab + ad + cd)^2 .

      =(ab + d (a+ c))^2 .

     =a^2b^2 + d^2 (a+ c)^2 + 2(ab)(d(a+c))

    =a^2b^2 + d^2 (a^2+ c^2+2ac) + 2a^2bd+2bcd

  Using equation (1) and (2),

  =a^2b^2 + 2a^2c^2+ 2b^2c^2+d^2a^2+2d^2b^2+d^2c^2

=a^2b^2 + a^2c^2+ a^2c^2+b^2c^2+b^2c^2+d^2a^2+d^2b^2+d^2b^2+d^2c^2

=a^2b^2 + a^2c^2+ a^2d^2+b^2.b^2+b^2c^2+b^2d^2+c^2b^2+c^2.c^2+d^2c^2

=a^2(b^2 + c^2+ d^2)+b^2(b^2+c^2+d^2)+c^2(b^2+c^2+d^2)

=(b^2 + c^2+ d^2)(a^2+b^2+c^2) = LHS

Hence proved

Question:26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer:

Let A, B be two numbers between 3 and 81 such that series    3, A, B,81  forms a GP.

Let a=first term and common ratio =r.

\therefore a_4=a.r^{4-1}

   81=3.r^{3}

27=r^{3}

 r=3

For r=3,

A=ar=(3)(3)=9

B=ar^2=(3)(3)^2=27

The, required numbers are 9,27.

Question:27 Find the value of n so that \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}  may be the geometric mean between a and b.

Answer:

M of a and b is \sqrt{ab}.

Given : 

          \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n}=\sqrt{ab}

Squaring both sides ,

     \left ( \frac{a^{n+1}+ b ^{n+1}}{a^n+b^n} \right )^2=ab

\left (a^{n+1}+ b ^{n+1})^2=({a^n+b^n} \right )^2ab

\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^2^n+b^2^n+2.a^n.b^n} \right )ab

\Rightarrow \left (a^{2n+2}+ b ^{2n+2}+2.a^{n+1}.b^{n+1})=({a^{2n+1}.b+a.b^{2n+1}+2.a^{n+1}.b^{n+1}} \right )

\Rightarrow \left (a^{2n+2}+ b ^{2n+2})=({a^{2n+1}.b+a.b^{2n+1}} \right )

\Rightarrow a^{2n+2}-{a^{2n+1}.b=a.b^{2n+1}} \right )- b ^{2n+2}

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

\Rightarrow a^{2n+1}=b^{2n+1}

\Rightarrow \left ( \frac{a}{b} \right )^{2n+1}=1

\Rightarrow \left ( \frac{a}{b} \right )^{2n+1}=1=\left ( \frac{a}{b} \right )^0

\Rightarrow 2n+1=0

\Rightarrow 2n=-1

\Rightarrow n=\frac{-1}{2}

Question:28 The sum of two numbers is 6 times their geometric mean, show that numbers  are in the ratio ( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 ) 

Answer:

Let there be two numbers a and b

geometric mean =\sqrt{ab}

According to the given condition,

a+b=6\sqrt{ab}

(a+b)^2=36(ab).............................................................(1)

Also,(a-b)^2=(a+b)^2-4ab=36ab-4ab=32ab

(a-b)=\sqrt{32}\sqrt{ab}

 (a-b)=4\sqrt{2}\sqrt{ab}.......................................................(2)

From (1) and (2), we get

2a=(6+4\sqrt{2})\sqrt{ab}

a=(3+2\sqrt{2})\sqrt{ab}

Putting the value of 'a' in (1),

b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}

b=(3-2\sqrt{2})\sqrt{ab}

\frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}

\frac{a}{b}=\frac{(3+2\sqrt{2})}{(3-2\sqrt{2})}

Thus, the ratio is ( 3+ 2 \sqrt 2 ) : ( 3 - 2 \sqrt 2 )

Question:29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A \pm \sqrt{( A+G)(A-G)}

Answer:

  If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.

AM=A=\frac{a+b}{2}

\Rightarrow a+b=2A...................................................................1

GM=G=\sqrt{ab}

\Rightarrow ab=G^2...........................................................................2

We know (a-b)^2=(a+b)^2-4ab

Put values from equation 1 and 2,

(a-b)^2=4A^2-4G^2

(a-b)^2=4(A^2-G^2)

(a-b)^2=4(A+G)(A-G)

(a-b)=4\sqrt{(A+G)(A-G)}..................................................................3

From 1 and 3 , we have

2a=2A+2\sqrt{(A+G)(A-G)}

\Rightarrow a=A+\sqrt{(A+G)(A-G)}

Put value of a in equation 1, we get

b=2A-A-\sqrt{(A+G)(A-G)}

\Rightarrow b=A-\sqrt{(A+G)(A-G)}

Thus, numbers are A \pm \sqrt{( A+G)(A-G)}

Question:30 The number of bacteria in a certain culture doubles every hour. If there were 30  bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?

Answer:

 The number of bacteria in a certain culture doubles every hour.It forms GP.

Given :      a=30   and  r=2.

a_3=a.r^{3-1}=30(2)^2=120

a_5=a.r^{5-1}=30(2)^4=480

a_n+_1=a.r^{n+1-1}=30(2)^n

Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and 30(2)^n respectively.

Question:31 What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Answer:

Given: Bank pays an annual interest rate of 10%  compounded annually.

Rs 500 amounts are deposited in the bank.

At the end of the first year, the amount 

                                                  =500\left ( 1+\frac{1}{10} \right )=500(1.1)

At the end of the second year, the amount =500(1.1)(1.1)

At the end of the third year, the amount =500(1.1)(1.1)(1.1)

At the end of 10 years, the amount =500(1.1)(1.1)(1.1)........(10times)

                                                =500(1.1)^{10}

Thus, at the end of 10 years, amount =Rs. 500(1.1)^{10}

Question:32 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation

Answer:

Let roots of the quadratic equation be a and b.

According to given condition,

               AM=\frac{a+b}{2}=8

                \Rightarrow (a+b)=16

                GM=\sqrt{ab}=5 

                 \Rightarrow ab=25

We know that   x^2-x(sum\, of\, roots)+(product\, of\, roots)=0

                   x^2-x(16)+(25)=0

                  x^2-16x+25=0

Thus, the quadratic equation = x^2-16x+25=0

 

NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.4

Question:1 Find the sum to n terms of each of the series in 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ...

Answer:

the series = 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ...

n th term  = n(n+1)=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)

                        =\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} k

                     =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}

                   =\frac{n(n+1)}{2}\left ( \frac{(2n+1)}{3}+1 \right )

                   =\frac{n(n+1)}{2}\left ( \frac{(2n+1+3)}{3} \right )

                  =\frac{n(n+1)}{2}\left ( \frac{(2n+4)}{3} \right )

                  =n(n+1)\left ( \frac{(n+2)}{3} \right )

                   = \frac{n(n+1)(n+2)}{3}

Question:2 Find the sum to n terms of each of the series in  1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...

Answer:

the series =  1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...

n th term  = n(n+1)(n+2)=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)(k+2)

                        =\sum _{k=1}^{n} k^3+3\sum _{k=1}^{n} k^2+2\sum _{k=1}^{n} k

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{3.n(n+1)(2n+1)}{6}+\frac{2.n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{2}+n(n+1)

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+(2n+1)+2)

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{n^2+n+4n+2+4}{2} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{n^2+5n+6}{2} \right )

                =\left [ \frac{n(n+1)}{4} \right ] \left ( n^2+5n+6 \right )

                 =\left [ \frac{n(n+1)}{4} \right ] \left ( n^2+2n+3n+6 \right )

                 =\left [ \frac{n(n+1)}{4} \right ] \left ( n(n+2)+3(n+2)\right )

                =\left [ \frac{n(n+1)}{4} \right ] \left ( (n+2)(n+3)\right )

              =\left [ \frac{n(n+1)(n+2)(n+3)}{4} \right ]

Thus, sum is 

                       =\left [ \frac{n(n+1)(n+2)(n+3)}{4} \right ]

Question:3 Find the sum to n terms of each of the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

Answer:

the series  3 \times 1 ^ 2 + 5 \times 2 ^ 2 + 7 \times +....+ 20 ^ 2

nth term  = (2n+1)(n^2)=2n^3+n^2=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} 2k^3+k^2

                        =2\sum _{k=1}^{n} k^3+\sum _{k=1}^{n} k^2

                     =2\left [ \frac{n(n+1)}{2} \right ]^2+\frac{n(n+1)(2n+1)}{6}

                    =\left [ \frac{n^2(n+1)^2}{2} \right ]+\frac{n(n+1)(2n+1)}{6}

                 =\left [ \frac{n(n+1)}{2} \right ](n(n+1)+\frac{(2n+1)}{3})

                   =\left [ \frac{n(n+1)}{2} \right ]\frac{(3n^2+3n+2n+1)}{3}

                   =\left [ \frac{n(n+1)}{2} \right ]\frac{(3n^2+5n+1)}{3}

                  = \frac{n(n+1)(3n^2+5n+1)}{6}   

Thus, the sum is 

                       = \frac{n(n+1)(3n^2+5n+1)}{6}

Question:4 Find the sum to n terms of each of the series in \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+ ...

Answer:

Series =

               \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+ ...

n^{th}\, term=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}

a_1=\frac{1}{1}-\frac{1}{2}

a_2=\frac{1}{2}-\frac{1}{3}

a_3=\frac{1}{3}-\frac{1}{4}.................................

a_n=\frac{1}{n}-\frac{1}{n+1}

a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} +\frac{1}{2}+\frac{1}{3}+............\frac{1}{n}\right ]-\left [ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.............\frac{1}{n+1} \right ]

a_1+a_2+a_3+...................a_n=\left [ \frac{1}{1} \right ]-\left [ \frac{1}{n+1} \right ]

                                         S_n=\frac{n+1-1}{n+1}

                                          S_n=\frac{n}{n+1}

Hence, the sum is 

                        S_n=\frac{n}{n+1}

Question:5 Find the sum to n terms of each of the series in 5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2

Answer:

series =    5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2

n th term  = (n+4)^2=n^2+8n+16=a_n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (k+4)^2

                        =\sum _{k=1}^{n} k^2+8\sum _{k=1}^{n} k+\sum _{k=1}^{n}16

                     =\frac{n(n+1)(2n+1)}{6}+\frac{8.n(n+1)}{2}+16n

           16th term is (16+4)^2=20^2

S_1_6=\frac{16(16+1)(2(16)+1)}{6}+\frac{8.(16)(16+1)}{2}+16(16)

          S_1_6=\frac{16(17)(33)}{6}+\frac{8.(16)(17)}{2}+16(16)

        S_1_6=1496+1088+256

    S_1_6=2840

Hence, the sum of the series 5 ^ 2 + 6 ^ 2 + 7 ^ 2 + ....+ 2 0 ^2  is 2840.

Question:6 Find the sum to n terms of each of the series 3 \times 8 + 6 \times 11 + 9\times 14+...

Answer:

series = 3 \times 8 + 6 \times 11 + 9\times 14+...

           =(n th term of 3,6,9,...........)\times(nth terms of 8,11,14,..........)  

n th term  = 3n(3n+5)=a_n=9n^2+15n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} 3k(3k+5)

                        =9\sum _{k=1}^{n} k^2+15\sum _{k=1}^{n} k

                     =\frac{9.n(n+1)(2n+1)}{6}+\frac{15.n(n+1)}{2}

                     =\frac{3.n(n+1)(2n+1)}{2}+\frac{15.n(n+1)}{2}  

                   =\frac{n(n+1)}{2}\left (3(2n+1)+15 \right )

                   =\frac{3.n(n+1)}{2}\left (2n+1+5 \right )

                  =\frac{3.n(n+1)}{2}\left (2n+6\right )

                =\frac{3.n(n+1)}{2}.2.\left (n+3\right )

                 =3.n(n+1)\left (n+3\right )

Hence, sum is  =3.n(n+1)\left (n+3\right )

Question:7 Find the sum to n terms of each of the series in 1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...

Answer:

series =    1 ^ 2 + ( 1 ^2 +2 ^ 2 ) + ( 1 ^ 2 +2 ^ 2 + 3 ^ 2 ) ...

n th term  = a_n=1^2+2^2+3^2+...................n^2=\frac{n(n+1)(2n+1)}{6}

                                                                                  =\frac{n(2n^2+3n+1)}{6}=\frac{2n^3+3n^2+n}{6}

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} \frac{2k^3+3k^2+k}{6}

                        =\frac{1}{3}\sum _{k=1}^{n} k^3+\frac{1}{2}\sum _{k=1}^{n} k^2+\frac{1}{6}\sum _{k=1}^{n} k

                     =\frac{1}{3}\left [ \frac{n(n+1)}{2} \right ]^2+\frac{1}{2}.\frac{n(n+1)(2n+1)}{6}+\frac{1}{6}\frac{n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{6} \right ] (\frac{n(n+1)}{2}+\frac{2n+1}{2}+\frac{1}{2})

                 =\left [ \frac{n(n+1)}{6} \right ] (\frac{n^2+n+2n+1+1}{2})

                  =\left [ \frac{n(n+1)}{6} \right ] (\frac{n^2+n+2n+2}{2})

            =\left [ \frac{n(n+1)}{6} \right ] (\frac{n(n+1)+2(n+1)}{2})

             =\left [ \frac{n(n+1)}{6} \right ] (\frac{(n+1)(n+2)}{2})

           =\left [ \frac{n(n+1)^2(n+2)}{12} \right ] 

Question:8 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n (n+1) ( n + 4 )

Answer:

nth terms is given by n (n+1) ( n + 4 )

a_n=n (n+1) ( n + 4 )=n(n^2+5n+4)=n^3+5n^2+4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k(k+1)(k+4)

                        =\sum _{k=1}^{n} k^3+5\sum _{k=1}^{n} k^2+4\sum _{k=1}^{n} k

                     =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+\frac{4.n(n+1)}{2}

                   =\left [ \frac{n(n+1)}{2} \right ]^2+\frac{5.n(n+1)(2n+1)}{6}+2.n(n+1)

                  =\left [ \frac{n(n+1)}{2} \right ] (\frac{n(n+1)}{2}+\frac{5(2n+1)}{3}+4)

                 =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+3n+20n+10+24}{6} \right )

                =\left [ \frac{n(n+1)}{2} \right ] \left ( \frac{3n^2+23n+34}{6} \right )

                =\left [ \frac{n(n+1)}{24} \right ] \left ( 3n^2+23n+34 \right )

Question:9 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by n^2 + 2 ^ n

Answer:

nth terms are given by n^2 + 2 ^ n

a_n=n^2 + 2 ^ n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} k^2+\sum _{k=1}^{n} 2^k

\sum _{k=1}^{n} 2^k=2^1+2^2+2^3+.....................2^n

This term is a GP with first term =a =2 and common ratio =r =2.

\sum _{k=1}^{n} 2^k=\frac{2(2^n-1)}{2-1}=2(2^n-1)

 

S_n=\sum _{k=1}^{n} k^2+2(2^n-1)

     S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)

Thus, the sum is 

                        S_n=\frac{n(n+1)(2n+1)}{6}+2(2^n-1)

Question:10 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by  ( 2n-1) ^2

Answer:

nth terms is given by  ( 2n-1) ^2.

a_n=( 2n-1) ^2=4n^2+1-4n

S_n=\sum _{k=1}^{n} a_k=\sum _{k=1}^{n} (2k-1)^2

                        =4\sum _{k=1}^{n} k^2-4\sum _{k=1}^{n} k+\sum _{k=1}^{n} 1

                     =\frac{4.n(n+1)(2n+1)}{6}-\frac{4.n(n+1)}{2}+n

                   =\frac{2.n(n+1)(2n+1)}{3}-2.n(n+1)+n

                    =n[\frac{2(n+1)(2n+1)}{3}-2(n+1)+1]

                   =n(\frac{4n^2+6n+2-6n-6+3}{3})

                     =n(\frac{4n^2-1}{3})

                     =n(\frac{(2n+1)(2n-1)}{3}) 

 

NCERT solutions for class 11 maths chapter 9 sequences and series-Miscellaneous Exercise

Question:1 Show that the sum of ( m+n)^{th} and ( m-n)^{th} terms of an A.P. is equal to twice the m^{th}term.

Answer:

Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

a_k=a+(k-1)d

\therefore a_m_+_n=a+(m+n-1)d

\therefore a_m_-_n=a+(m-n-1)d

a_m=a+(m-1)d

a_m_+_n+ a_m_-_n=a+(m+n-1)d+a+(m-n-1)d

                           =2a+(m+n-1+m-n-1)d

                         =2a+(2m-2)d

                       =2(a+(m-1)d)

                        =2.a_m

Hence, the sum of ( m+n)^{th} and ( m-n)^{th} terms of an A.P. is equal to twice the m^{th}term.