NCERT Solutions Class 11 Maths Chapter-9 (Sequences and Series)Miscellaneous Exercise

NCERT Solutions Class 11 Maths from class 11th Students will get the answers of Chapter-9 (Sequences and Series)Miscellaneous Exercise . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Solutions Class 11 Maths Chapter-9 (Sequences and Series)Miscellaneous Exercise

NCERT Question-Answer

Class 11 Mathematics

Chapter-9 (Sequences and Series)Miscellaneous Exercise

Questions and answers given in practice

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 mth term.
Ans:

Question 2:
If the sum of three numbers in AP is 24 and their product is 440, find the numbers.
Ans:
Let the three numbers in A.P. be a – d, a, and a + d.
According to the given information,
(a – d) + (a) + (a + d) = 24
3a = 24
⇒ a = 8
and (a – d) a (a + d) = 440 ………………(ii)
(8 – d) (8) (8 + d) = 440
(8 – d) (8 + d) = 55
= 64 – d2 = 55
d2 = 64 – 55 = 9
d = ±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.

Question 3:
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:

Hence, S3 = 3 (S2 – S1)

Question 4:
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer. 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 $\begin{aligned} ∴ S_{29} = \frac{29}{2} (203 + 399) \\ = \frac{29}{2} (602) \\ = (29) (301) \\ = 8729 \\ Thus, the required sum is 8729 \end{aligned}$

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

Question 6:
Find the sum of all two digit numbers which when divided by 4, yields 1 as remninder.
Ans:
The sum of two digit numbers divisible by 4 yield 1 as remainder is 13 + 17 + 21 + ………… + 97.
Let the sum be denoted by S and let 97 be the nth term.
∴ Tn = a + (n – 1) d
97 = a + (n – 1) d
= 13 + (n – 1) 4
⇒ 97 = 13 + 4n – 4
⇒ 97 – 9 = 4n
⇒ n = 22
∴ The sum, Sn = 13 + 17 + 21 + ………….+ 97
∴ Sn = n2 [2a + (n – 1)d]
= 227 [2 × 13 + (22 – 1) × 4]
= 11 [26 + 21 × 4]
= 11 [26 + 84]
= 11 × 110 = 1210

Question 7:

Ans:
It is given that,
f(x + y) = f(x) × f(y) for all x, y ∈ N
f(1) = 3
Taking x = y = 1 in eq. (i), we obtain
f(1 + 1) = f(2) = f(1)
f(1) = 3 × 3 = 9
Similarly, f(1 + 1 + 1) = f(3) = f(1 + 2) = f(1) f(2) = 3 × 9 = 27
f(4) = f(1 + 3) = f(1) f(3) = 3 × 27 = 81
f(1), f(2), f(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, Sn = a(rn−1)r−1
It is given that, ∑nx=1 f(x) = 120
∴ 120 = 3(3n−1)3−1

Question 8:
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, Sn = a(rn−1)r−1
It is given that the first term a is 5 and common ratio r is 2.
315 = 5(2n−1)2−1
⇒ 2n = 64 = (2)6
⇒ n = 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.

Question 9:
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:

Question 10:
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 a, ar and ar2.
From the given condition, a + ar + ar2 = 56
a (1 + r + r2) = 56
a = 561+r+r2 …………………(i)
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
a(r – 1)2 = 8 ………….(ii)
561+r+r2 (r – 1)2 = 8 [using eq. (1)]
⇒ 7 (r2 – 2r + 1) = 1 + r + r2
⇒ 7r2 – 14r + 7 – 1 – r – r2 = 0
⇒ 6r2 – 15r + 6 = 0
⇒ r2 – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3)(r – 2) = 0
∴ r = 2, 12
When r = 2, a = 8; When r = 12, a = 32
Therefore, when r = 2, the three numbers in G.P. are 8, 16 and 32.
When r = 12, the three numbers in G.P. are 32, 16 and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.

Question 11:
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:

Question 12:
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 a, a + d, a + 2d, a + 3d,… a + (n – 2) d, a + (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.

Question 13. If then show that a,b,c and d are in G.P.

Ans:

Question 14. 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:

Proved.
Question 15: The pth,qthand rth terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0

Ans:

Question16. If are in A.P., prove that a , b ,c are in A.P

Ans:

Question 17. If a, b, c, d are in G.P, prove that (an + bn),(bn + cn),(cn + dn) are in G.P.

Ans:
Given: a, b, c, d are in G.P.
b2 = ac
c2 = bd
ad = bc
To prove:
(an + bn),(bn + cn),(cn + dn)
are in G.P.
That is, (bn + cn)2 = (an + bn),(cn + dn)
Then,
L.H.S =(bn + cn)2
=b2n+2bncn+c2n
=(b2)n+2bncn+(c2)n
=(ac)n+2bncn+(bd)n
=ancn+bncn+bncn+bndn
=ancn+bncn+andn+bndn
=cn(an+bn)+dn(an+bn)
=(an+bn)(an+dn)
=
R.H.S
Therefore,
(bn+cn)2=(an+bn)(cn+dn)
Therefore, (bn+cn),(bn+cn)
and (cn+dn)
are in G.P.
Question 18. If a and b are the roots of x2 − 3x + p = 0 and c, d are roots of x2−12x + q= 0 where a,b,c,d form a G.P.Prove that (q+p) : (q−p)= 17:15.

Ans:
Given: a
and b
are the roots of x2 − 3x + p =0
.Therefore,
a + b = 3 and ab=p …(1)
We also know that c and d
are the roots of x2−12x+q=0
.Therefore,
c+d=12 and cd=q …(2)
Also, a,b,c,d are in G.P.
Let us take a=x, b=xr, c=xr2
and d=xr3
We get from (1) and (2) that,
x+xr = 3
⇒ x(1+r) =3Also,xr2 + xr3 = 12
⇒xr2 + (1+r) = 12

Therefore, Therefore, it is proved that (q+p):(q−p) = 17:15as we obtain the same for both the cases.
Question 19. The ratio of the A.M. and G.M. of two positive numbers a and b is m : n Show that a : b =

Ans:
Given: a : b =

By component and dividend,

Question 20. If a, b, c are in A.P.; b, c, d are in G.P. and 1/c,1/d,1/e are in A.P. prove that a, c, e are in G.P.

Ans:
Since,

Given a,b,c are in A.P.Therefore, b−a=c−b are in A.P.
⇒ 2b = a + c

Question 21. Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + ……….
(ii) .6 + .66 + .666 + ………….

Ans:
(i) sn = 5 + 55 + 555 + ………. up to n terms
= 5/9 [1 + 11 + 111 + ………. up to n terms]

(ii) sn = .6 + .66 + .666 + …………. up to n terms
= 6 [.1 + .11 + .111 + ………. up to n terms]

Question 22. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.

Ans:
Given: 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms
Therefore the nth term an=2n×(2n+2)=4n2+4n
Then,
a20 =4(20)2 + 4(20)
=4(400)+80
=1600+80
=1680
Therefore, 1680
is the 20th
Question23. Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 21 + ……….
Ans:
Given: 3 + 7 + 13 + 21 + 21 + ……………….(i)
Also 3+7+13+21+31+...
is the given series.
S=3+7+13+21+31+...+an−1 + an
S=3+7+13+21+...+ an−2 + an−1 + an ……….(ii)
Subtracting eq. (ii) from eq. (i),

Question 24. If S1,S2,S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22= S3 (1+8S1)

Ans:

Question 25. Find the sum of the following series up to n terms:

Ans:
Given:

up to nth terms

Question 26. Show that

Ans:
Given:

Question 27. A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual installments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

Ans:
It is given that Rs.6000
is paid in cash by the farmer.
Therefore, the unpaid amount is given by
Rs.12000−Rs.6000 = Rs.6000
According to the conditions given in the question, the interest to be paid annually by the farmer is
12 of 6000 , 12 of 5500 , 12 of 5000...12 of 500
Therefore, the total interest to be paid by the farmer
=12 of 6000+12 of 5500+12 of 5000+...+12 of 500
=12 of (6000+5500+5000+...+500)
=12 of (500+1000+1500+...+6000)
With both the first term and common difference equal to 500
, the series 500,1000,1500...6000 is an A.P.
Let n
be the number of terms of the A.P.
Therefore,
6000=500+(n−1)500
⇒1+(n−1)=12
⇒n=12
Therefore, the sum of the given A.P.
=122[2(500)+(12−1)(500)]
=6[1000+5500]
=6(6500)
=39000
Therefore, the total interest to be paid by the farmer
=12 of (500+1000+1500+...+6000)
=12 of Rs.39000
= Rs.4680
Therefore, the total cost of tractor
= (Rs.12000+ Rs.4680 )= Rs.16680
Therefore, the total cost of the tractor is Rs.16680
Question 28. Shams had Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Ans:
It is given that for Rs.22000 Shamshad Ali buys a scooter and Rs.4000
is paid in cash. Therefore, the unpaid amount is given by
Rs.22000−Rs.4000 = Rs.18000
According to the conditions given in the question, the interest to be paid annuallyis
10 of 18000 , 10 of 17000, 10 of 16000...10 of 1000
Therefore, the total interest to be paid by the farmer
=10 of 18000+10 of 17000+1 of 16000+...+10 of 1000
=10 of (18000+17000+16000+...+1000)
=10 of (1000+2000+3000+...+18000)
With both the first term and common difference equal to 1000
, the series 1000,2000,3000...18000 is an A.P.
Let n be the number of terms of the A.P.
Therefore,
18000=1000+(n−1)1000
⇒1+(n−1)=18
⇒n=18
Therefore, the sum of the given A.P.
=18/2 [2(1000)+(18−1)(1000)]
=9[2000+17000]
=9(19000)
=171000
Therefore, the total interest to be paid
=10 of (18000+17000+16000+...+1000)
=10 of Rs.171000=Rs.17100
Therefore, the total cost of scooter=
(Rs.22000 + Rs.17100)=Rs.39100
Therefore, the total cost of the scooter is Rs.39100
Question 29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

Ans:

=

Question 30. A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

Ans:
Rs.10000 is deposited by the man in a bank at the rate of 5 simple interest annually
=5/ 100 × Rs.10000 = Rs.500
Therefore,
10000+500+500+...+500 is the interest in 15th
year. (500 is 14 added times)
Therefore, the amount in 15th
year = Rs.10000+14 × Rs.500
= Rs.10000 + Rs.7000= Rs.17000
Rs.10000+500+500+...+500
is the amount after 20 years. (500 is 20 added times)
Therefore, the amount after 20
years = Rs.10000+20 × Rs.500
= Rs.10000 + Rs.10000=Rs.20000
The total amount after 20 years is Rs.20000
Question 31. A manufacturer reckons that the value of a machine, which cost him Rs. 15625 will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Ans:
The cost of the machine is Rs.15625.
Every year machine depreciates by 20.
Therefore, 80 of the original cost ,i.e., 45 of the original cost is its value after every year.
Therefore, the value at the end of 5 years
=15626×45×45×...×45
=5×1024
=5120
Therefore, Rs.5120
is the value of the machine at the end of 5
years.
Question 32. 150 workers were engaged to finish a job in a certain number of boys. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work find the number of days in which the work was completed.

Ans:
Let the number of days in which 150 workers finish the work be x
.According to the conditions given in the question,
150x=150+146+142+...(x+8)
terms
With first term a=146, common difference d=−4 and number of turns as (x+8) , the series 150+146+142+...(x+8)
terms is an A.P.
⇒150x=(x+8)/2[2(150)+(x+8−1)(−4)]
⇒150x=(x+8)[150+(x+7)(−2)]
⇒150x=(x+8)(150−2x−14)
⇒150x=(x+8)(136−2x)
⇒75x=(x+8)(68−x)
⇒75x=68x−x2+544−8x
⇒x2+75x−60x−544=0
⇒x2+15x−544=0
⇒x2+32x−17x−544=0
⇒x(x+32)−17(x+32)=0
⇒(x−17)(x+32)=0
⇒x=17
or x=−32
We know that x cannot be negative.
So, x=17
Therefore, 17
is the number of days in which the work was completed. Then the required number of days =(17+8)=25