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Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.6 | Set 3
  • Last Updated : 07 Apr, 2021

Question 49. Find the sum of first n odd natural numbers.

Solution:

First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. 

First term(a) = 1, common difference(d) = 3 – 1 = 2 

and nth term(an) = 2n – 1.

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

= n[1 + 2n – 1] / 2



= 2n2 / 2

= n2

Hence, the sum of first n odd natural numbers is n2

Question 50. Find the sum of 

(i) all odd numbers between 0 and 50.

Solution:

All odd numbers between 0 and 50 are 1, 3, 5, 7, . . . 49. 

first term(a) = 1, common difference(d) = 3 – 1 = 2 

and nth term(an) = 49.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 49 = 1 + (n – 1)2



=> 2(n – 1) = 48

=> n – 1 = 24

=> n = 25

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S25 = 25[1 + 49] / 2

= 25[25]

= 625

Hence, the sum of all odd numbers between 0 and 50 is 625.

(ii) all odd numbers between 100 and 200.

Solution:

All odd numbers between 100 and 200 are 101,103,105,107, . . . 199. 

First term(a) = 101, 

common difference(d) = 103 – 101 = 2 

and nth term(an) = 199.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 199 = 101 + (n – 1)2

=> 2(n – 1) = 98

=> n – 1 = 49

=> n = 50

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S50 = 50[101 + 199]/2

= 25[300]

= 7500

Hence, the sum of all odd numbers between 100 and 200 is 7500.

Question 51. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Solution:

Odd integers between 1 and 1000 which are divisible by 3 are 3, 9, 15, . . . .999.

First term(a) = 3, common difference(d) = 9 – 3 = 6 

and nth term(an) = 999.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 999 = 3 + (n – 1)6

=> 6(n – 1) = 996

=> n – 1 = 166

=> n = 167

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S167 = 167[3 + 999] / 2 

= 167[501]

= 83667

Hence Proved.

Question 52. Find the sum of all integers between 84 and 719, which are multiples of 5. 

Solution:

Integers between 84 and 719, which are multiples of 5 are 85, 90, 95, 100, . . . . 715.

First term(a) = 85, common difference(d) = 90 – 85 = 5 and nth term(an) = 715.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 715 = 85 + (n – 1)5

=> 5(n – 1) = 630

=> n – 1 = 126

=> n = 127

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S127 = 127[85 + 715] / 2

= 127[400]

= 50800

Hence, the sum of all integers between 84 and 719, which are multiples of 5 is 50800.

Question 53. Find the sum of all integers between 50 and 500 which are divisible by 7.

 Solution:

All integers between 50 and 500 which are divisible by 7 are 56, 63, 70, 77, . . . . 497.

First term(a) = 56, common difference(d) = 63 – 56 = 7 and nth term(an) = 497.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 497 = 56 + (n – 1)7

=> 7(n – 1) = 441

=> n – 1 = 63

=> n = 64

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

 S64 = 64[56 + 497] / 2

= 64[553]

= 17696

Hence, the sum of all integers between 50 and 500 which are divisible by 7 is 17696.

Question 54. Find the sum of all even integers between 101 and 999.

Solution:

All even integers between 101 and 999 are 102, 104, 106, 108, . . . . 998.

First term(a) = 102, common difference(d) = 104 – 102 = 2 and nth term(an) = 998.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 998 = 102 + (n – 1)2

=> 2(n – 1) = 896

=> n – 1 = 448

=> n = 449

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S449 = 449[102 + 998] / 2

= 449[550]

= 246950

Hence, the sum of all even integers between 101 and 999 is 246950.

Question 55. Find the sum of all integers

(i) between 100 and 550 which are divisible by 9

Solution:

All integers between 100 and 550 which are divisible by 9 are 108, 117, 126, 135, . . . .549.

First term(a) = 108, common difference(d) = 117 – 108 = 9 and nth term(an) = 549.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 549 = 108 + (n – 1)9

=> 9(n – 1) = 441

=> n – 1 = 49

=> n = 50

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S50 = 50[108 + 549]/2

= 25[657]

= 16425

Hence, sum of all integers between 100 and 550 which are divisible by 9 is 16425.

(ii) between 100 and 550 which are not divisible by 9

Solution:

All integers between 100 and 550 which are divisible by 9 are 108, 117, 126, 135, . . . .549.

First term(a) = 108, common difference(d) = 117 – 108 = 9 and nth term(an) = 549.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 549 = 108 + (n – 1)9

=> 9(n – 1) = 441

=> n – 1 = 49

=> n = 50

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S50 = 50[108 + 549] / 2

= 25[657]

= 16425 

Now, we know,

Sum of integers between 100 and 550 which are not divisible by 9 = Sum of integers between 100 and 550 – Sum of integers between 100 and 550 which are divisible by 9

= [101+102+103+104+. . . . .+549] –  S50

= [1+2+3+4+. . . . +549] – [1+2+3+4+. . . .+100] – 16425

= 549[550]/2 –  100[101]/2 – 16425   

= 150975 – 5050 – 16425

= 129500

Hence, sum of all integers between 100 and 550 which are not divisible by 9 is 129500.

(iii) between 1 and 500 which are multiples of 2 as well as of 5.

Solution:

All integers between 1 and 500 which are multiples of 2 as well as of 5 are 10, 20, 30, 40, . . . .490.

First term(a) = 10, common difference(d) = 20 – 10 = 10 and nth term(an) = 490.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 490 = 10 + (n – 1)10

=> 10(n – 1) = 480

=> n – 1 = 48

=> n = 49

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S49 = 49[10 + 490] / 2

= 49[250]

= 12250

Hence, sum of all integers between 1 and 500 which are multiples of 2 as well as of 5 is 12250.

(iv) from 1 to 500 which are multiples of 2 as well as of 5.

Solution:

All integers from 1 and 500 which are multiples of 2 as well as of 5 are 10, 20, 30, 40, . . . .500.

First term(a) = 10, common difference(d) = 20 – 10 = 10 and nth term(an) = 500.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 500 = 10 + (n – 1)10

=> 10(n – 1) = 490

=> n – 1 = 49

=> n = 50

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S50 = 50[10 + 500]/2

= 25[510]

= 12750

Hence, sum of all integers from 1 to 500 which are multiples of 2 as well as of 5 is 12750.

(v) from 1 to 500 which are multiples of 2 or 5.

Solution:

Integers from 1 to 500 which are multiples of 2 are 2, 4, 6, 8, . . . .500.

First term(a) = 2, common difference(d) = 4 – 2 = 2 and nth term(an) = 500.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

500 = 2 + (n – 1)2

=> 2(n–1) = 498

=> n–1 = 249

=> n = 250

Let S1 be the sum of this A.P. Hence, S1 = 250[2 + 500] / 2 = 125[502] = 62750.

Integers from 1 to 500 which are multiples of 5 are 5, 10, 15, 20, . . . .500.

First term(a) = 5, common difference(d) = 10 – 5 = 5 and nth term(an) = 500.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

500 = 5 + (n – 1)5

=> (n – 1)5 = 495

=> n – 1 = 99

=> n = 100

Let S2 be the sum of this A.P. Hence, S2 = 100[5 + 500] / 2 = 50[505] = 25250.

Integers from 1 to 500 which are multiples of 2 as well as 5 are 10, 20, 30 . . . .500.

We know, 500 = 10 + (n – 1)10

=> 10(n – 1) = 490

=> n – 1 = 49

=> n = 50

Let S3 be the sum of this A.P. Hence, S3 = 50[10 + 500] / 2 = 25[510] = 12750.

Hence, required sum = S1 + S2 – S3

= 62750 + 25250 – 12750

= 75250

Hence, sum of all integers from 1 to 500 which are multiples of 2 or 5 is 75250.

Question 56. Let there be an A.P. with first term a and common difference d. If an denotes its nth term and Sn is the sum of first n terms, then find:

(i) n and Sn, if a = 5, d = 3 and an = 50.

Solution:

Given A.P. has a = 5, d = 3 and an = 50.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 50 = 5 + (n – 1)3

=> 3(n – 1) = 45

=> n – 1 = 15

=> n = 16 

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

S16 = 16[5 + 50] / 2

= 8[55]

= 440

Hence, the value of n is 16 and sum is 440.

(ii) n and a, if an = 4, d = 2 and Sn = –14

Solution:

Given A.P. has d = 2, an = 4 and Sn = –14. 

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 4 = a + (n – 1)2

=> 4 = a + 2n – 2

=> a = 6 – 2n  . . . . (1)

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

=> –14 = n[a + 4] / 2

=> n[a + 4] = –28

=> n[6 – 2n + 4] = –28   [Using eq(1)]

=> 10n – 2n2 = –28

=> n2 – 5n – 14 = 0

=> n2 – 7n + 2n – 14 = 0

=> n(n – 7) + 2(n – 2) = 0

=> (n – 7) (n + 2) = 0

=> n = 7 or n = –2

Ignoring n = –2 as number of terms cannot be negative. So, we get n = 7.

On putting n = 7 in (1), we get, a = 6 – 2(7) = –8

Hence, the value of n is 7 and a is –8.

(iii) d, if a = 3, n = 8 and Sn = 192

Solution:

Given A.P. has a = 3, n = 8 and Sn = 192.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

=> 192 = 8[2(3) + (8 – 1)d] / 2

=> 4[6+7d] = 192

=> 6+7d = 48

=> 7d = 42

=> d = 6

Hence, the value of d is 6.

(iv) a if an = 28, Sn = 144 and n = 9

Solution:

 Given A.P. has an = 28, Sn = 144 and n = 9.

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

=> 144 = 9[a + 28] / 2

=> a + 28 = 32

=> a = 4

Hence, the value of a is 4.

(v) n and d if a = 8, an = 62 and Sn = 210

Solution:

 Given A.P. has a = 8, an = 62 and Sn = 210.

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So, 

=> 210 = n[8 + 62] / 2

=> 70n = 420

=> n = 6

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 62 = 8 + (6 – 1)d

=> 5d = 54

=> d = 54/5

Hence, the value of n is 6 and d is 54/5.

(vi) n and an, if a = 2, d = 8 and Sn = 90.

Solution:

 Given A.P. has a=2, d=8 and Sn=90.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

=> 90 = n[2(2) + (n – 1)8] / 2

=> 90 = n[2 + 4n – 4)]

=> 4n2 – 2n – 90 = 0

=> 4n2 – 20n + 18n – 90 = 0

=> 4n(n – 5) + 18n(n – 5) = 0

=> (n – 5) (4n + 18) = 0

=> n = 5 or n = –9/2

Ignoring n = –9/2 as n cannot be a fraction as well as negative. So, we get n = 5.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

= 2 + (5 – 1)8

= 2 + 32 

= 34

Hence, the value of n is 5 and an is 34.

(vii) k, if Sn = 3n2 + 5n and ak = 164.

Solution:

Given Sn = 3n2 + 5n,

On putting n = 1, we get the first term(a), S1 = a = 3(1)2 + 5(1) = 8

On putting n = 2 gives S2 = a + a + d = 3(2)2 + 5(2) = 22

=> d = 22 – 2a

=> d = 22–16 = 6

The kth term of the A.P., ak = a + (k – 1)d = 164

=> 8 + (k – 1)6 = 164

=> 6k = 162

=> k = 27

Hence, the value of k is 27.

(viii) S22, if d = 22 and a22 = 149

Solution:

We know, a22 = a + 21d = 149 . . . . (1)

On putting d = 22 (given) in eq(1), we get

=> a + 21(22) = 149

=> a = 149 – 462

=> a = –313

We know, S22 = 22[a + a22] / 2

= 22[–313 + 149] / 2

= 11[164]

= 1804

Hence, the value of sum is 1804.

Question 57. If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 – S4).

Solution:

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

Therefore, L.H.S. = S12 = 12[2a + (12 – 1)d] / 2

= 6[12a + 11d]

= 12a + 66d

R.H.S. = 3(S8 – S4)

= 3[8(2a + (8 – 1)d) / 2 – 4(2a + (4 – 1)d) / 2]

= 3[4(2a + 7d) – 2(2a + 3d)]

= 3[8a + 28d – 4a – 6d]

= 3[4a + 22d]

= 12a + 66d

= L.H.S.

Hence proved.

Question 58. A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

Solution:

Suppose n minutes is the time taken by the policeman to catch the thief.

Since policeman started running 2 minutes later, the thief ran for (n + 2) minutes.

Therefore, distance travelled by thief = Speed × Time = 50(n + 2) meter.

Speed of the police after every minute are: 60, 65, 70,. . . . 

These form an A.P. with first term(a) = 60 and common difference(d) = 65 – 60 = 5

Total distance travelled by police in n minutes = n[2(60) + (n – 1)5] / 2

= n[120 + 5n – 5] / 2

= n[115 + 5n] / 2

Now, according to the question,

Distance travelled by thief in (n + 2) minutes = Distance travelled by police in n minutes

=> 50(n + 2) = n[115 + 5n] / 2

=> 100(n + 2) = 115n +5n2

=> 5n2 + 15n – 200 = 0

=> n2 + 3n + 40 = 0

=> n2 + 8n – 5n + 40 = 0

=> n(n + 8) – 5(n + 8)

=> n = 5 or n = – 8

Ignoring n = – 8 as time cannot be negative. So, we get n = 5.

Hence, after 5 minutes the policeman will catch the thief.

Question 59. The sums of first n terms of three A.P.s are S1, S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2.

Solution:

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

According to the question,

S1 = n[2(5)+(n – 1)2] / 2 = n[8 + 2n] / 2 = 4n + n2

S2 = n[2(5) + (n – 1)4] / 2 = n[6 + 4n] / 2 = 3n + 2n2

S3 = n[2(5) + (n – 1)6] / 2 = n[4 + 6n] / 2 = 2n + 3n2

Now, L.H.S. = (4n + n2) + (2n + 3n2

= 6n + 4n2

= 2[3n + 2n2]

= 2S2

Hence proved.

Question 60. Resham wanted to save at least Rs. 6500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her saying by Rs.20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?

Solution:

Given that Resham has saved in following sequence in a year (every month record): 450, 470, 490,. . . .

This sequence is an A.P. with first term(a) = 450 and common difference(d) = 20.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

Now, sum of the savings after 12 months would be S12 as value of n in this case is 12.

S12 = 12[2(450) + (12 – 1)20] / 2

= 6[900 + 220]

= 6[1120]

= 6720

So, Resham would save Rs 6720 by the end of 12 months which is greater than Rs 6500, 

she would be able to send her daughter to the school next year.

Question 61. In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.

Solution:

Trees planted by the student in a class A = 2(A) [Since there are 2 sections for every class.]

Trees planted by the students from class 1 to 12 in sequence can be written as:

4, 8, 12,………., 48

First term(a) = 4 and common difference, d = 8 − 4 = 4

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So, 

=> 48 = 4 + (n − 1)4

=> 4(n − 1) = 44

=> n − 1 = 11

=> n = 12

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

According to the question,

Number of trees planted = Sum of the A.P. series = S12 

= 12[4+48]/2 = 6[52] = 312

Hence, 312 trees were planted by the students.

Question 62. Ramkali would need Rs. 1800 for admission fee and books etc., for her daughter to start going to school from the next year. She saved Rs. 50 in the first month of this year and increased her monthly saving by Rs. 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school?

Solution:

Given that Ramkali has saved in following sequence in a year(every month record): 50, 70, 90,. . . .

First term(a) = 50 and common difference(d) = 20.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

Now, sum of the savings after 1 year(12 months) would be S12 as value of n in this case is 12.

S12 = 12[2(50) + (12 – 1)20] / 2

= 6[100 + 220]

= 6[320]

= 1920

Hence, Resham would save Rs 1920 by the end of 1 year(12 months) which is 

greater than Rs 1800, she would be able to fulfill her dream of sending her daughter to school.

Question 63. A man saved Rs 16500 in ten years. In each year after the first year, he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?

Solution:

So, the common difference(d) = 100 and sum of savings done in 10 years(S10) = 16500

We have to find the savings in first year, i.e., first term(a). 

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So,

=> S10 = 16500 

=> 10[2a + (10 – 1)100] / 2 = 16500

=> 5[2a + 900] = 16500

=> 10a = 16500 – 4500

=> 10a = 12000

=> a = 1200

Hence, value of savings in first year is 1200.

Question 64. A man saved Rs 32 during the first year, Rs 36 during the second year and in this way he increases his savings by Rs 4 per year. In what time his savings would be Rs 200?

Solution:

So, the first term(a) = 32 and common difference(d) = 4.

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

Here, Sn = 200

So, 

=> n[2(32) + (n – 1)4] / 2 = 200

=> n[64 + 4n – 4] / 2 = 200

=> n[60 + 4n] / 2 = 200

=> 2n2 + 30n – 200 = 0

=> n2 + 15n – 100 = 0

=> n2 + 20n – 5n – 100 = 0

=> n(n + 20) – 5(n + 20) = 0

=> (n – 5) (n + 20) = 0

=> n = 5 or n = –20

Ignoring n = –20 as number of terms cannot be negative. So, we get n = 5.

Therefore, in 5 years his savings would be Rs 200.

Question 65. A man arranges to pay off his debt of Rs 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first installment.

Solution: 

Given that the installments are forming an arithmetic series. 

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

Now, Amount of 40 installments = Rs 3600

=> S40 = 3600

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

=> 40[2a + (40 – 1)d] / 2 = 3600

=> 2a + 39d = 180  . . . . (1)

Amount of 30 installments = Rs 3600 – Rs [3600 / 3] = 3600 – 1200 = 2400

 => S30 = 2400

=> 30[2a + (30 – 1)d] / 2 = 2400

=> 2a + 29d = 160  . . . . (2)

On subtracting eq(2) from (1), we get,

=> (2a + 39d) – (2a + 29d) = 180 – 160

=> 10d = 20

=> d = 2

On putting d = 2 in (1), we get,

=> 2a + 39(2) = 180

=> 2a = 180–78

=> 2a = 102

=> a = 51

Hence, value of first installment is 51.

Question 66. There are 25 trees at equal distances of 5 meter in a line with a well, the distance of the well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Solution:

Given number of trees(n) = 25. 

Distance covered by the gardener during watering of first tree = 2(10) = 20 meters

Distance covered by the gardener during watering of second tree = 2(10 + 5) = 30 meters

Distance covered by the gardener during watering of third tree = 2(10 + 5 + 5) = 40 meters

So, first term(a) = 20 and common difference(d) = 30 – 20 = 10 meters.

Total distance covered to water 25 trees = Sum of 25 terms of A.P. = S25

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

we get

= 25[2(20) + (25 – 1)10] / 2 

= 25[40 + 240] / 2

= 25[140]

= 3500 meters

Hence, the total distance the gardener will cover in order to water all the trees is 3500 meters.

Question 67. A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this, he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

Solution:

 Total amount to be counted = Rs 10710.

Amount man would count at the rate of Rs 180 per minute for 1/2 hour = 180(30) = Rs 5400

Now amount left before the rate starts decreasing = 10710–5400 = Rs 5310

This amount of 5310 is counted at a rate of Rs 3 less every minute than the preceding minute. 

So, first term(a) = 5310/30 = 177, common difference = –3 and Sn = 5310

Now by using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

We get

=> n[2(177) + (n – 1)(–3)] / 2 = 5310

=> n[357 – 3n] = 10620

=> 3n2 – 357n + 10620 = 0

=> n2 – 119n + 3540 = 0

=> n2 – 60n – 59n + 3540 = 0

=> n(n – 60) – 59(n – 60) = 0

=> n = 60 or n = 59

Ignoring n = 60 as n cannot be equal to or greater than 60. So, we get n = 59.

So, total time taken = 30+59 = 89 minutes.

Hence, time taken by him to count the entire amount is 89 minutes.

Question 68. A piece of equipment cost a certain factory Rs 600,000. If it depreciates in value, 15% the first, 13.5% the second year, 12% the third year and so on. What will be its value at the end of 10 years, all percentages applying to the original cost? 

Solution:

Cost of equipment = Rs 600,000

Depreciation value in first year = 15% of 600,000 = 90,000

Depreciation value in second year = 13.5% of 600000 = 81,000

Depreciation value in third year = 12% of 600000 = 72,000

So, first term(a) = 90000 and common difference(d) = 81000 – 90000 = –9000.

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

Total Depreciation amount in 10 years = S10

= 10[2(90000) + (10 – 1)(–9000)] / 2 

= 5[180000 – 81000]

= 5[99000]

= Rs 495,000

Value of equipment = Cost – Depreciation at end of 10 years

= 600000 – 495000

= Rs 105000

Hence, value at the end of 10 years is Rs 105000.

Question 69. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize. 

Solution:

Total amount that has to be given(S7) = Rs 700

Number of Prizes(n) = 7

As each prize is Rs 20 less than its preceding prize, 

these prizes are forming an A.P. with common difference(d) = –20

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

=> 700 = 7[2(a) + (7 – 1)(–20)] / 2

=> 2a – 120 = 200

=> 2a = 320

=> a = 160

The value of 1st prize = 160

The value of 2nd prize = 160 – 20 = Rs 140 

The value of 3rd prize = 140 – 20 = Rs 120

The value of 4th prize = 120 – 20 = Rs 100

The value of 5th prize = 100 – 20 = Rs 80

The value of 6th prize = 80 – 20 = Rs 60

The value of 7th prize = 60 – 20 = Rs 40

Question 70. If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 – S10).

Solution:

Let’s take L.H.S.

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

We get

S30 = 30[2a + (30 – 1)d] / 2

= 15[2a + 29d]

R.H.S. = 3(S20 – S10).

= 3[20[2a + (20 – 1)d] / 2 – 10[2a + (10 – 1)d] / 2]

= 3[10(2a + 19d) – 5(2a + 9d)]

= 3[20a + 190d – 10a – 45d]

= 3[10a + 145d]

= 3 × 5[2a + 29d]

= 15[2a + 29d]

= S30

Hence proved.

Question 71. Solve the equation (-4) + (-1) + 2 + 5 + …. + x = 437.

Solution:

So, first term(a) = –4, common difference(d) = –1 – (–4) = 3, last term(an) = x and sum(Sn) = 437.

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

We get

=> 437 = n[2(–4) + (n – 1)3] / 2

=> 874 = n[–8 + 3n – 3]

=> 3n2 – 11n – 874 = 0

=> 3n2 – 57n + 46n – 874 = 0

=> 3n(n – 19) + 46(n – 19) = 0

=> n = 19 or n = –46/3

Ignoring n = –46/3 as n cannot be a fraction as well as negative. So, we get n =19.

Now, we know nth term of an A.P. is given by an = a + (n – 1)d.

=>  x = –4 + (19 – 1)3

=> x = –4 + 54

=> x = 50

Hence, the value of x is 50.

Question 72. Which term of the A.P. -2, -7, -12,…, will be -77? Find the sum of this A.P. up to the term -77.

Solution:

Given A.P. has first term(a) = –2, common difference(d) = –7 – (–2) = –5, and given term(an) = –77

Now, we know nth term of an A.P. is given by an = a + (n – 1)d.

=> –77 = –2 + (n – 1)(–5)

=> –5(n – 1) = –75 

=> n – 1 = 15

=> n = 16

By using the formula of the sum of n terms of an A.P.

Sn = n[2a +(n – 1)d] / 2

So, sum up to the term –77 = S16

= 16[2(–2) + (16 – 1)(–5)]/2

= 8[–4 – 75]

= 8[–79]

= –632

Hence, 16th term of the A.P. will be –77 and sum up to this term is –632.

Question 73. The sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.

Solution:

Let us assume S1 be the sum of the first n terms of an A.P. 

whose first term is 8 and the common difference is 20.

Hence, S1 = n[2(8) + (n – 1)20] / 2

= n[16 + 20n – 20] / 2

= n[10n – 2]

Let us assume S2 be the sum of first 2n terms of another A.P. 

whose first term is -30 and common difference is 8.

Hence, S2 = 2n[2(–30) + (2n – 1)8] / 2

= n[–60 + 16n – 8]

= n[16n – 68]

According to the question, we have

=> S1 = S2

=> n[10n – 2] = n[16n – 68]

=> 10n – 2 = 16n – 68

=> 6n = 66

=> n = 11

Hence, the value of n is 11.

Question 74. The students of a school decided to beautify the school on the annual day by fixing colorful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 meter. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag? 

Solution:

Ruchi has to fix 13 flags to the left of the middle position, 1 flag at the middle position and

 remaining 26 flags at the right of the middle position.

Distance covered in fixing 1st flag to the left of the middle position = 2 + 2 = 4m

Distance covered in fixing 2nd flag to the left of the middle position = 4 + 4 = 8m

Distance covered in fixing 3rd flag to the left of the middle position = 8 + 8 = 16m

So, first term(a) = 4, common difference(d) = 8 – 4 = 4, number of terms = 13 

and last term(a13) = 26+26 = 52m 

So we use the sum of n terms of an A.P.: Sn = n[2a + (n – 1)d] / 2.

So, Distance covered in fixing flags to the left of the middle position = Sum of 13 terms of the A.P.

= 13[2(4) + (13 – 1)4] / 2

= 13[8 + 48]/2

= 13[28]

= 364

Total distance covered = 2 × 364 = 728m

Hence, the maximum distance Ruchi travelled carrying a flag 

would be the distance she travelled while carrying the last flag 

to the left or to the right, which in our case is 26m.

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