Skip to content
Related Articles

Related Articles

Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.2
  • Last Updated : 28 Dec, 2020

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

Solution:

The odd integers forms an A.P. with 

Common difference(d)=2  and  first term (a)=1

Here, a+(n-1)d=2001

⇒1+(n-1)2=2001



⇒2n-2=2000

⇒n=1001

Sn = n[2a + (n – 1)d]/2

S1001=1001[2(1)+(1001-1)2]/2

=1001[2+2000]/2

=1001[2002]/2

=1001[1001]

=1002001

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

Question 2. In an A.P. the first term is 2 and the sum of the first 5 terms is one-fourth of the next 5 terms. Show that the 20th term is -112.

Solution:

The first term(a)=2.

Let the common difference be d.

Sum of first 5 terms=10+10d.

Sum of next 5 terms=10+35d.

According to given condition ,

10+10d=¼(10+35d)

⇒40+40d=10+35d

⇒5d=-30



⇒d=-6

a20=2+(20-1)(-6)=2-(19)6=2-114=-112

Hence, proved

Question 3. If the sum of first n terms of an A.P. is (pn + qn2), where p and q are constants. Find the common difference.

Solution:

We know that, 

Sn = n[2a + (n – 1)d]/2

According to the given condition,

n[2a+(n-1)d]/2 = pn+qn2

⇒n[2a+nd-d]/2 = pn+qn2

⇒na+n2(d/2)-n(d/2) = pn+qn2

Comparing the coefficients of n2 on both sides we get,

d/2=q ⇒ d=2q

Therefore, the common difference is 2q.

Question 4. How many terms of the A.P. -6, -11/7, -5,… are needed to give the sum -25?

Solution:

Let the number of terms to give the sum -25 be n.

We know S​​​​​​n = n(2a+(n-1)d)/2 where n is the number of terms, a is the first term and d is the common difference.

Here, a=-6 and d=(-11/2)+6=(-11+12)/2=1/2.

Substituting these values in the formula of sum we get,

-25=n[2(-6)+(n-1)(1/2)]/2

⇒-50=n[-12+(n/2)-(1/2)]

⇒-50=n[(-25/2)+(n/2)]

⇒-100=n(-25+n)

⇒n2-25n+100=0

⇒n2-5n-20n+100=0

⇒n(n-5)-20(n-5)=0

⇒n=20 or 5

Therefore, the number of terms required is 5 or 20.

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

Solution:

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

Given,

pth term=ap =a+(p-1)d=1/q               —(1)

qth term = aq= a+(q-1)d = 1/p           —(2)

Subtracting (2) from (1):

(p-1)d-(q-1)d=(1/q)-(1/p)

⇒(p-1-q+1)d=(p-q)/pq

⇒(p-q)d=(p-q)/pq

⇒d=1/pq

Putting value of d in (1):

a+(p-1)/pq=1/q

⇒a=(1/q)-(1/q)+(1/pq)=1/pq

Therefore,

Spq = pq[2a+(pq-1)d]/2 

= pq[(2/pq)+((pq-1)/pq)]/2 

=1+(pq-1)/2

=(pq+1)/2

The sum of the first pq terms of the A.P. is (pq+1)/2.

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

Solution:

Let the number of terms required to give the sum of 116 be n

We know, Sn=n[2a+(n-1)d]/2

Given, a=25 and d= 22-25=-3

Substituting the values we get,

116=n[2(25)+(n-1)(-3)]/2

⇒232=n[50-3n+3]

⇒232=n(53-3n)=53n-3n2

⇒3n2-53n+232=0

⇒3n2-24n-29n+232=0

⇒3n(n-8)-29(n-8)=0

⇒(n-8)(3n-29)=0

⇒n=8 or 29/3

But, n cannot be 29/3.Therefore, n=8.

a8 is the last term=a+(n-1)d=25+(8-1)(-3)=25+(7)(-3)=25-21=4

Therefore, the last term is 4.

Question 7. Find the sum of n terms of A.P. if it’s kth term is given as 5k+1.

Solution:

Given, kth term is 5k+1

kth term = ak = a+(k-1)d

a+(k-1)d=5k+1

⇒ a + kd – d=5k+1

Comparing the coefficients on both sides, we get d=5

a-d=1⇒a-5=1⇒a=6

Sn=n[2a+(n-1)d]/2 

= n[2(6)-(n-1)5]/2 

=n[12+5n-5]/2 

= n(5n+7)/2.

Question 8. Find the sum of all natural numbers between 100 and 1000, which are multiples of 5.

Solution:

The natural numbers between 100 and 1000 which are multiples of 5 are, 105, 110, . . . , 995. It can be seen as an A.P. starting with a= 105 and common difference d=5.

a+(n-1)d=995

⇒105+(n-1)5=995

⇒(n-1)5=995-105=890

⇒n-1=178

⇒n=179

Sn=179[2(105)+(179-1)5]/2

=179[2(105)+(178)5]/2

=179[105+(89)5]

=179[105+445]

=179[550]

=98450

The required sum is 98450.

Question 9. The sum of n terms of two A.P. are in the ratio 5n+4:9n+6. Find the ratio of their 18th terms.

Solution:

Let a1, a2 and d1, d2 be the first term and the common difference of the first and the second A.P. respectively.

Given,

(Sum of n terms of first A.P)/(Sum of n terms of second A.P.)=(5n+4)/(9n+6)

⇒{n[2a1-(n-1)d1]/2}/{n[2a2+(n-1)d2]/2}=(5n+4)/(9n+6)

⇒[2a1+(n-1)d1]/[2a2+(n-1)d2]=(5n+4)/(9n+6)                                                         —(1)

Substituting n=35 in (1), 

(2a1+34d1)/(2a2+34d2)=[5(35)+4]/[9(35)+6]

⇒[a1+17d1]/[a2+17d2]=179/321                                                                             —(2)

(18th term of the first A.P)/(18th term of the second A.P.)=(a1+17d1)/(a2+17d2)    —(3)

R.H.S. of (3) =L.H.S. of (2). So, we get,

(18th term of the first A.P)/(18th term of the second A.P.)=179/321

Therefore, the ratio of the 18th term of both the A.P. 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.

Solution:

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

Given,

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

⇒p[2a+(p-1)d]=q[2a+(q-1)d]

⇒2ap+pd(p-1)=2aq+qd(q-1)

⇒2a(p-q)+d[p(p-1)-q(q-1)]=0

⇒2a(p-q)+d[p2-p-q2+q]=0

⇒2a(p-q)+d[(p-q)(p+q)-(p-q)]=0

⇒2a(p-q)+d[(p-q)(p+q-1)]=0

⇒2a+d(p+q-1)=0

⇒d=-2a/(p+q-1)                                         —(1)

So, Sp+q =(p+q)[2a+(p+q-1)d]/2

Putting value of d ,

⇒Sp+q=(p+q)[2a-2a]/2=0

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

Question 11. Sum of first p, q, and r terms of an A.P. are a, b and c respectively. Prove that {a(q-r)/p}+{b(r-p)/q}+{c(p-q)/r=0.

Solution:

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

Given,

Sp=p[2a1+(p-1)d]/2=a

⇒2a1=(p-1)d=2a/p                                                      —(1)

Sq=q[2a1+(q-1)d]/2=b

⇒2a1+(q-1)d=2b/q                                                      —(2)

Sr=r[2a1+(r-1)d]/2=c

⇒2a1+(r-1)d=2c/r                                                         —(3)

Subtracting (2) from (1), we get

(p-1)d-(q-1)d=(2a/p)-(2b/q)

⇒d(p-1-q+1)=(2aq-2bq)/pq

⇒d=2(aq-bp)/pq(p-q)                                                    —(4)

Subtracting (3) from (2),we get

(q-1)d-(r-1)d=(2b/q)-(2c/r)

⇒d(q-1-r+1)=(2b/q)-(2c/r)

⇒d(q-r)=(2br-2qc)/qr

⇒d={2(br-qc)}/{qr(q-r)}                                                  —(5)

Equating both values of d in (4) and (5),we obtain

(aq-bp)/(pq(p-q))=(br-qc)/(qr(q-r))

⇒qr(q-r)(aq-bq)=pq(p-q)(br-qc)

⇒r(aq-bp)(q-r)=p(br-qc)(p-q)

⇒(aqr-bpr)(q-r)=(bpr-pqc)(p-q)

Dividing both sides by pqr,

{(a/p)-(b/q)}(q-r)={(b/q)-(c/r)}(p-q)

⇒(a/p)(q-r)-(b/q)(q-r+p-q)+(c/r)(p-q)=0

⇒(a(q-r)/p)+(b(r-p)/q)+(c(p-q)/r)=0. 

Hence, proved

Question 12. The ratio of sum of m and n terms of A.P. is m2:n2. Show that the ratio of mth and nth term is (2m-1):(2n-1).

Solution:

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

Given,

(Sum of m terms)/(Sum of n terms)=m2/n2

⇒[m(2a+(m-1)d)/2]/[n(2a+(n-1)d)/2]=m2/n2

⇒[2a+(m-1)d]/[2a+(n-1)d]=m/n                                              —(1)

Putting m=2m-1 and n=2n-1 in (1), we get

[2a+(2m-2)d]/[2a+(2n-2)d]=(2m-1)/(2n-1)

⇒[a+(m-1)d]/[a+(n-1)d]=[2m-1]/[2n-1]                                   —(2)

[mth term of A.P.]/[nth term of A.P.]=[a+(m-1)d]/[a+(n-1)d]      —(3)

R.H.S. of (3) = L.H.S. of (2). So, we get,

[mth term of A.P.]/[nth term of A.P.]=(2m-1)/(2n-1) 

Hence, proved

Question 13. If the sum of n terms of an A.P. is 3n2+5n and it’s mth term is 164, find the value of m.

Solution:

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

am =a+(m-1)d=164                                                                   —(1)

Sn =n[2a+(n-1)d]/2

Given,

n[2a+nd-d]/2=3n2 +5n

⇒na+(n2d/2)=3n2 +5n

Comparing coefficients on both sides,

d/2=3⇒d=6    and a-(d/2)=5⇒a-3=5 ⇒ a=8

From (1),

8+(m-1)6=164

⇒ (m-1)6=164-8=156

⇒ m-1=26

⇒ m=27

Therefore, the value of m is 27.

Question 14. Insert 5 numbers between 8 and 26 such that the resulting sequence is in A.P.

Solution:

Let A1, A2, A3, A4 and A5 be 5 numbers between 8 and 26 such that it forms an A.P.

a=8, b=26, n=7

We can see, 26=8+(7-1)d

⇒6d=26-8=18

⇒d=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+4d=8+4(3)=8+12=20

A5=a+5d=8+5(3)=8+15=23

Therefore, the required 5 numbers between 8 and 26 to form an A.P. are 11, 14, 17, 20 and 23.

Question 15. If  (an + bn )/(an-1 + bn-1) is the A.M. between a and b. Find the value of n.

Solution:

A.M. of a and b=(a+b)/2

According to given,

(a+b)/2=(an +bn )/(an-1 +bn-1)

⇒(a+b)(an-1+bn-1)=2((an+bn)

⇒an+abn-1+ban-1+bn=2an+2bn

⇒abn-1+an-1b=an+bn

⇒abn-1-bn=an-an-1b

⇒bn-1(a-b)=an-1(a-b)

⇒bn-1=an-1

⇒(a/b)n-1=1=(a/b)0

⇒n-1=0

n=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 7th and (m-1)th term is 5:9. Find the value of m.

Solution:

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

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

31=1+(m+2-1)d

⇒30=(m+1)d

⇒d=30/(m+1)                                                                —(1)

A1=a+d

A2=a+2d

A3=a+3d

A4=a+4d . . .

A7=a+7d

Am=a+(m-1)d

According to given,

(a-7d)/[a+(m-1)d]=5/9

⇒[1+7(30/(m+1))]/[1+(m-1)(30/m+1)]=5/9             [From (1)]

⇒[m+1+7(30)]/[m+1+30(m-1)]=5/9

⇒[m+1+210]/[m+1+30m-30]=5/9

⇒[m+211]/[31m-29]=5/9

⇒9m+1899=155m-145

⇒155m-9m=1899+145

⇒146m=2044

⇒m=14

Therefore, the value of m is 14.

Question 17.  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?

Solution:

Given, he first installment of the loan is Rs 100. The second installment of the loan in Rs 105 and so on. We can see that the amount that the man pays every month forms an A.P.

First term, a=100. Common difference, d=5

a30 =a+(30-1)d

=100+(29)5

=100+145

=245

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

Question 18. The angles of a polygon will form an A.P. with common difference d as 5° and first term a as 120°. Find the number of sides of the polygon.

Solution:

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

That is, Sn =180°(n-2)

⇒n[2a+(n-1)d]/2=180°(n-2)

⇒n[240+(n-1)5]=360(n-2)

⇒240n+5n2 -5n=360n-720

⇒5n2+235n-360n+720=0

⇒5n2-125n+144=0

⇒n2-16n-9n+144=0

⇒n(n-16)-9(n-16)=0

⇒(n-9)(n-16)=0

n = 9 or 16

My Personal Notes arrow_drop_up
Recommended Articles
Page :