RD SHARMA SOLUTION CHAPTER- 9 Arithmetic Progressions | CLASS 10TH MATHEMATICS-EDUGROWN

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.1

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Write the first five terms for the sequence whose nth terms is:

a_n = n² – n + 1Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 1 (ix)

Solution 1 (ix)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Solution 2 (iii)

Question 2 (iv)

Find the indicated terms in the sequence whose nth terms are :

a_n = (n – 1) (2 – n) (3 + n); a₁, a₂, a₃Solution 2 (iv)

Question 2 (v)

Find the indicated terms in the sequence whose nth terms are:

a_n = (-1)ⁿ n; a₃, a₅, a₈Solution 2 (v)

Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 3 (iv)

Solution 3 (iv)

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5

Solution 5

Question 6(i)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 2n – 1Solution 6(i)

Question 6(ii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 3n² + 5Solution 6(ii)

Question 6(iii)

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_n = 1 + n + n²Solution 6(iii)

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.3

Question 1

Solution 1

Question 2

Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = -3

(ii) a = -1, d = 1/2

(iii) a = -1.5, d = -0.5Solution 2

Question 3 (i)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.


Solution 3 (i)

Question 3 (ii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.Solution 3 (ii)

Question 3(iii)

In which of the following situations, the sequence of numbers formed will form an A.P.?

Divya deposited Rs.1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …., and so on.Solution 3(iii)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Solution 5 (iv)

Question 5 (v)

Solution 5 (v)

Question 5 (vi)

Find out whether of the given sequence is an arithemtic progressions. If it is an arithmetic progressions, find out the common difference.

p, p + 90, p + 180, p + 270, … where p = (999)⁹⁹⁹Solution 5 (vi)

Question 5 (vii)

Solution 5 (vii)

Question 5 (viii)

Solution 5 (viii)

Question 5 (ix)

Solution 5 (ix)

Question 5 (x)

Solution 5 (x)

Question 5 (xi)

Solution 5 (xi)

Question 5 (xii)

Solution 5 (xii)

Question 6

Solution 6

Question 7

Solution 7

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.4

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 2 (i)

Solution 2 (i)

Question 2 (ii)

Solution 2 (ii)

Question 2 (iii)

Which term of the A.P. 4, 9, 14, … is 254?Solution 2 (iii)

Question 2 (iv)

Solution 2 (iv)

Question 2 (v)

Solution 2 (v)

Question 2 (vi)

Which term of the A.P. -7, -12, -17, -22,… will be -82? Is -100 any term of the A.P.?Solution 2 (vi)

The given A.P. is -7, -12, -17, -22,…

First term (a) = -7

Common difference = -12 – (-7) = -5

Suppose nth term of the A.P. is -82.

So, -82 is the 16^th term of the A.P.

To check whether -100 is any term of the A.P., take a_n as -100.

So, n is not a natural number.

Hence, -100 is not the term of this A.P.Question 3 (i)

Solution 3 (i)

Question 3 (ii)

Solution 3 (ii)

Question 3 (iii)

Solution 3 (iii)

Question 4 (i)

Solution 4 (i)

Question 4 (ii)

Solution 4 (ii)

Question 4 (iii)

Solution 4 (iii)

Question 4 (iv)

Solution 4 (iv)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9 (i)

Solution 9 (i)

Question 9 (ii)

Find the 12^th term from the end of the following arithmetic progressions:

3, 8, 13, …, 253Solution 9 (ii)

A.P. is 3, 8, 13, …, 253

We have:

Last term (l) = 253

Common difference (d) = 8 – 3 = 5

Therefore,

12^th term from end

       = l – (n – 1)d

       = 253 – (12 – 1) (5)

       = 253 – 55

       = 198Question 9 (iii)

Solution 9 (iii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The 26^th, 11^th and last term of an A.P. are 0, 3 and , respectively. Find the common difference and the number of terms.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 34. Find the first term and the common difference of the A.P.Solution 18

Question 19

Solution 19

Question 20 (i)

Solution 20 (i)

Question 20 (ii)

Solution 20 (ii)

Question 21

Solution 21

Question 22 (i)

Solution 22 (i)

Question 22 (ii)

Solution 22 (ii)

Question 22 (iii)

Solution 22 (iii)

Question 22 (iv)

Solution 22 (iv)

Question 23

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15^th term.Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

The 17^th term of an A.P. is 5 more than twice its 8^th term. If the 11^th term of the A.P. is 43, find the n^th term.Solution 37

Thus, n^th term is given by

a_n = a + (n – 1)d
a_n = 3 + (n – 1)4
a_n = 3 + 4n – 4
a_n = 4n – 1Question 38

Find the number of all three digit natural numbers which are divisible by 9.Solution 38

The smallest three digit number divisible by 9 = 108

The largest three digit number divisible by 9 = 999

Here let us write the series in this form,

108, 117, 126, …………….., 999

a = 108, d = 9

t_n = a + (n – 1)d

999= 108 + (n – 1)9

⇒ 999 – 108 = (n – 1)9

⇒ 891 = (n – 1)9

⇒ (n – 1) = 99

⇒ n = 99 + 1

∴ n = 100

Number of terms divisible by 9

Number of all three digit natural numbers divisible by 9 is 100.Question 39

The 19^th term of an A.P. is equal to three times its sixth term. If its 9^th term is 19, find the A.P.Solution 39

Question 40

The 9^th term of an A.P. is equal to 6 times its second term. If its 5^th term is 22, find the A.P.Solution 40

Question 41

The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is 4 times its 15^th term.Solution 41

Let the first term be ‘a’ and the common difference be ‘d’

t₂₄ = a + (24 – 1)d = a + 23d

t_10­ = a + (10 – 1)d = a + 9d

t₇₂ = a + (72 – 1)d = a + 71d

t₁₅ = a + (15 – 1)d = a + 14d

t₂₄ = 2t₁₀

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 23d – 18d = 2a – a

∴ 5d = a

t₇₂ = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t₁₅

∴t₇₂ = 4t₁₅Question 42

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.Solution 42

L.C.M. of 2 and 5 = 10

3- digit number after 100 divisible by 10 = 110

3- digit number before 999 divisible by 10 = 990

Let the number of natural numbers be ‘n’

990 = 110 + (n – 1)d

⇒ 990 – 110 = (n – 1) × 10

⇒ 880 = 10 × (n – 1)

⇒ n – 1 = 88

∴ n = 89

The number of natural numbers between 110 and 999 which are divisible by 2 and 5 is 89.Question 43

If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)^rd term.Solution 43

Let the first term be ‘a’ and the common difference be ‘d’.

Question 44

The sum of 5^th and 9^th terms of an A.P. is 30. If its 25^th term is three times its 8^th term, find the A.P.Solution 44

Question 45

Find where 0(zero) is a term of the A.P. 40, 37, 34, 31, …Solution 45

Let the first term be ‘a’ and the common difference be ‘d’.

a = 40

d = 37 – 40 = – 3

Let the n^th term of the series be 0.

t_n = a + (n – 1)d

⇒ 0 = 40 + (n – 1)( – 3)

⇒ 0 = 40 – 3(n – 1)

⇒ 3(n – 1) = 40

∴ No term of the series is 0.Question 46

Find the middle term of the A.P. 213, 205, 197, …, 37.Solution 46

Given A.P. is 213, 205, 197, …, 37.

Here, first term = a = 213

And, common difference = d = 205 – 213 = -8

a_n = 37

n^th term of an A.P. is given by

a_­n = a + (n – 1)d

⇒ 37 = 213 + (n – 1)(-8)

⇒ 37 = 213 – 8n + 8

⇒ 37 = 221 – 8n

⇒ 8n = 221 – 37

⇒ 8n = 184

⇒ n = 23

So, there are 23 terms in the given A.P.

⇒ The middle term is 12^th term.

⇒ a₁₂ = 213 + (12 – 1)(-8)

= 213 + (11)(-8)

= 213 – 88

= 125

Hence, the middle term is 125.Question 47

If the 5^th term of an A.P. is 31 and 25^th term is 140 more than the 5^th term, find the A.P.Solution 47

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

Then, we have

a₅ = 31 and a₂₅ = a₅ + 140

⇒ a + 4d = 31 and a + 24d = a + 4d + 140

⇒ a + 4d = 31 and 20d = 140

⇒ a + 4d = 31 and d = 7

⇒ a + 4(7) = 31 and d = 7

⇒ a + 28 = 31 and d = 7

⇒ a = 3 and d = 7

Hence, the A.P. is a, a + d, a + 2d, a + 3d, ……

i.e. 3, 3 + 7, 3 + 2(7), 3 + 3(7), ……

i.e. 3, 10, 17, 24, …..Question 48

Find the sum of two middle terms of the A.P.

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?Solution 51

Question 52

Find the 12^th term from the end of the A.P. -2, -4, -6, …, -100.Solution 52

Question 53

For the A.P.: -3, -7, -11, …, can we find a₃₀ – a₂₀ without actually finding a₃₀ and a₂₀? Give reason for your answer.Solution 53

Question 54

Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10^th terms is the same as the difference between their 21^st terms, which is the same as the difference between any two corresponding terms. Why?Solution 54

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.Solution 8

Question 9

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.Solution 9

Let the first three terms of an A.P. be a – d, a, a + d

As per the question,

(a – d) + a + (a + d) = 18

∴ 3a = 18

∴ a = 6

Also, (a – d)(a + d) = 5d

∴ (6 – d)(6 + d) = 5d

∴ 36 – d² = 5d

∴ d² + 5d – 36 = 0

∴ d² + 9d – 4d – 36 = 0

∴ (d + 9)(d – 4) = 0

∴ d = -9 or d = 4

Thus, the terms will be 15, 6, -3 or 2, 6, 10.Question 10

Spilt 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.Solution 10

Question 11

The angles of a triangle are in A.P. the greatest angle is twice the least. Find all the angles.Solution 11

Question 12

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.Solution 12

Chapter 9 – Arithmetic Progressions Exercise Ex. 9.6

Question 1 (i)

Solution 1 (i)

Question 1 (ii)

Solution 1 (ii)

Question 1 (iii)

Solution 1 (iii)

Question 1 (iv)

Solution 1 (iv)

Question 1 (v)

Solution 1 (v)

Question 1 (vi)

Solution 1 (vi)

Question 1 (vii)

Solution 1 (vii)

Question 1 (viii)

Solution 1 (viii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find the sum of last ten terms of the A.P: 8, 10, 12, 14,.., 126.Solution 4

Question 5 (i)

Solution 5 (i)

Question 5 (ii)

Solution 5 (ii)

Question 5 (iii)

Solution 5 (iii)

Question 5 (iv)

Find the sum of the first 15 terms of each of the following sequences having n^th term as

y_n = 9 – 5nSolution 5 (iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10 (i)

Solution 10 (i)

Question 10 (ii)

How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?Solution 10 (ii)

Question 10 (iii)

How many terms of the A.P. 9, 17, 25, … must be taken so that their sum is 636?

Remark* – Question modified.Solution 10 (iii)

Question 10 (iv)

Solution 10 (iv)

Question 10(v)

How many terms of the A.P. 27, 24, 21… should be taken so that their sum is zero?Solution 10(v)

Question 10 (vi)

How many terms of the A.P. 45, 39, 33 … must be taken so that their sum is 180? Explain the double answer.Solution 10 (vi)

Let the required number of terms be n.

As the given A.P. is 45, 39, 33 …

Here, a = 45 and d = 39 – 45 = -6 

The sum is given as 180

∴ S_n = 180

When n = 10,

When n = 6,

Hence, number of terms can be 6 or 10.Question 11 (i)

Solution 11 (i)

Question 11 (ii)

Solution 11 (ii)

Question 11 (iii)

Solution 11 (iii)

Question 12(i)

Find the sum of

The first 15 multiples of 8.Solution 12(i)

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

Question 12(ii)

Find the sum of

The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.Solution 12(ii)

a) divisible by 3 3,6,9,… Now, n=40, a=3, d=3

b) divisible by 5 5,10,15,… Now, n=40, a=5, d=5

c)divisible by 6 6,12,18,… Now, n=40, a=3, d=6Question 12(iii)

Find the sum of

All 3 – digit natural numbers which are divisible by 13.Solution 12(iii)

Three-digit numbers divisible by 13 are 104,117, 130,…988. Now, a=104, l=988Question 12(iv)

Find the sum of

All 3 – digit natural numbers, which are multiples of 11. Solution 12(iv)

Three-digit numbers which are multiples of 11 are 110,121, 132,…990. Now, a=110, l=990Question 12(v)

Find the sum of

All 2 – digit natural numbers divisible by 4. Solution 12(v)

Two-digit numbers divisible by 4 are 12,16,…96. Now, a=12, l=96Question 12 (vi)

Find the sum of first 8 multiples of 3.Solution 12 (vi)

The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, …

These are in A.P. with,

first term (a) = 3 and common difference (d) = 3

To find S₈ when a = 3, d = 3

Hence, the sum of first 8 multiples of 3 is 108.Question 13 (i)

Solution 13 (i)

Question 13 (ii)

Solution 13 (ii)

Question 13 (iii)

Solution 13 (iii)

Question 13 (iv)

Solution 13 (iv)

Question 13 (v)

Solution 13 (v)

Question 13 (vi)

Solution 13 (vi)

Question 13 (vii)

Solution 13 (vii)

Question 13 (viii)

Find the sum:

Solution 13 (viii)

Let ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22 (i)

Solution 22 (i)

Question 22 (ii)

If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.Solution 22 (ii)

Sum of first n terms of an AP is given by

As per the question, S₄ = 40 and S₁₄ = 280

Also,  

Subtracting (i) from (ii), we get, 10d = 20

Therefore, d = 2

Substituting d in (i), we get, a = 7

Sum of first n terms becomes

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.Solution 26

Let the number of terms be ‘n’, ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

⇒ 49 = 7 + (n – 1)d

⇒ 42 = (n – 1)d…..(i)

∴ 840 = n[14 + (n – 1)d]……(ii)

Substituting (ii) in (i),

840 = n[14 + 42]

⇒ 840 = 56n

∴ n = 15

Substituting n in (i)

42 = (15 – 1)d

Common difference, d =3Question 27

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.Solution 27

Let the number of terms be ‘n’, ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

⇒ 45 = 5 + (n – 1)d

⇒ 40 = (n – 1)d…..(i)

∴ 800 = n[10 + (n – 1)d]……(ii)

Substituting (ii) in (i),

800 = n[10 + 40]

⇒ 800 = 50n

∴ n = 16

Substituting n in (i)

40 = (16 – 1)d

Question 28

The sum of first q terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1 : 2. Find the first and 15^th term of the A.P.Solution 28

Question 29

If the 10^th term of an A.P. is 21 and the sum of its first ten terms is 120, find its n^th term.Solution 29

Let ‘a’ be the first term and ‘d’ be the common difference.

t_n = a + (n – 1)d

t₁₀ = a + (10 – 1)d

⇒ 21 = a + 9d……(i)

120 = 5[2a + 9d]

24 = 2a + 9d………(ii)

(ii) – (i) ⇒ 

a = 3

Substituting a in (i), we get

a + 9d = 21

⇒ 3 + 9d = 21

⇒ 9d = 18

∴d = 2

t_n = a + (n – 1)d

= 3 + (n – 1)2

= 3 + 2n – 2

∴ t_n= 2n + 1Question 30

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^th term of this A.P.Solution 30

Let the first term be ‘a’ and the common difference be ‘d’  

63 = 7[a + 3d]

9 = a + 3d……….(i)

Sum of the next 7 terms = 161

Sum of the first 14 terms = 63 + 161 = 224

224 = 7[2a + 13d]

32 = 2a + 13d………..(ii)

Solving (i) and (ii), we get

d = 2, a = 3

28^th term of the A.P., t₂₈ = a + (28 – 1)d

= 3 + 27 × 2

= 3 + 54

= 57

∴ The 28^th of the A.P. is 57.Question 31

The sum of first seven terms of an A.P. is 182. If its 4^th and the 17^th terms are in the ratio 1: 5, find the A.P.Solution 31

Let the first term be ‘a’ and the common difference be ‘d’.

26 × 2 = [2a + (7 – 1)d]

52 = 2a + 6d

26 = a + 3d……..(i)

From (i) and (ii),

⇒ 13d = 104

∴d = 8

From (i), a = 2

The A.P. is 2, 10, 18, 26,…….Question 32

The n^th term of an A.P. is given by (- 4n + 15). Find the sum of the first 20 terms of this A.P.Solution 32

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

t_n = – 4n + 15

t₁ = – 4 × 1 + 15 = 11

t₂ = – 4 × 2 + 15 = 7

t₃ = – 4 × 3 + 15 = 3

Common Difference, d = 7 – 11 = -4

= 10 × (-54)

= – 540

*Note: Answer given in the book is incorrect.Question 33

Solution 33

Question 34

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^th term.Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Find the number of terms of the A.P. – 12, – 9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.Solution 37

Let the number of the terms be ‘n’.

Common Difference, d = – 9 + 12 = 3

t_n = a + (n – 1)d

⇒ 21 = – 12 + 3(n – 1)

⇒ 21 + 12 = 3(n – 1)

⇒ 3(n – 1) = 33

⇒ n – 1 = 11

∴n = 12

Number of terms of the series = 12

If 1 is added to each term of the above A.P.,

– 11, – 8, – 5,…….,22

Number of terms in the series, n₁ = 12

Sum of all the terms,

The sum of the terms = 66Question 38

The sum of the first n terms of an A.P. is 3n² + 6n. Find the nth term of this A.P.Solution 38

Sum of n terms of the A.P., S_n = 3n² + 6n

S₁ = 3 × 1² + 6 × 1 = 9 = t₁ ……(i)

S₂ = 3 × 2² + 6 × 2 = 24 = t₁ + t₂ …….(ii)

S₃ = 3 × 3² + 6 × 3 = 45 = t₁ + t₂ + t₃ ……..(iii)

From (i), (ii) and (iii),

t₁ = 9, t₂ = 15, t₃ = 21

Common difference, d = 15 – 9 = 6

n^th of the AP, t_n = a + (n – 1)d

= 9 + (n – 1) 6

= 9 + 6n – 6

= 6n + 3

Thus, the n^th term of the given A.P. = 6n + 3Question 39

The sum of the first n terms of an A.P. is 5n – n². Find the n^th term of this A.P.Solution 39

S_n = 5n – n²

S₁ = 5 × 1 – 1² = 4 = t₁………..(i)

S₂ = 5 × 2 – 2² = 6 = t₁ +t₂………..(ii)

S₃ = 5 × 3 – 3² = 6 = t₁ + t₂ + t_3­……….(iii)

From (i), (ii) and (iii),

t₁ = 4, t₂ = 2, t₃ = 0

Here a = 4, d = 2 – 4 = – 2

t_n = a + (n – 1)d

= 4 + (n – 1)( -2)

= 4 – 2n + 2

= 6 – 2nQuestion 40

The sum of the first n terms of an A.P. is 4n² + 2n. find the n^th term of this A.P.Solution 40

S_n = 4n² + 2n

S₁ = 4 × 1² + 2 × 1 = 6 = t₁………….(i)

S₂ = 4 × 2² + 2 × 2 = 20 = t₁ + t₂……….(ii)

S₃ = 4 × 3² + 2 × 3 = 42 = t₁ + t₂ + t₃………..(iii)

From (i), (ii) and (iii),

t₁ = 6, t₂ = 14, t₃ = 22

Here a = 6, d = 14 – 6 = 8

t_n = a + (n – 1)d

t_n = 6 + (n – 1)8

= 6 + 8n – 8

= 8n – 2

*Note: Answer given in the book is incorrect.Question 41

The sum of first n terms of an A.P. is 3n² + 4n. find the 25^th term of this A.P.Solution 41

Sum of n terms of the A.P., S_n = 3n² + 4n

S₁ = 3 × 1² + 4 × 1 = 7 = t₁………(i)

S₂ = 3 × 2² + 4 × 2 = 20 = t₁ + t_2­…….(ii)

S₃ = 3 × 3² + 4 × 3 = 39 = t₁ + t₂ + t₃ …….(iii)

From (i), (ii), (iii)

t₁ = 7, t₂ = 13, t­₃ = 19

Common difference, d = 13 – 7 = 6

25^th of the term of this A.P., t₂₅ = 7 + (25 – 1)6

= 7 + 144 = 151

∴The 25^th term of the A.P. is 151.Question 42

The sum of first n terms of an A.P. is 5n² + 3n. If its m^th term is 168, find the value of m. Also, find the 20^th term of this A.P.Solution 42

Sum of the terms, S_n = 5n² + 3n

S₁ = 5 × 1² + 3 × 1 = 8 = t₁………..(i)

S₂ = 5 × 2² + 3 × 2 = 26 = t₁ + t₂…………..(ii)

S₃ = 5 × 3² + 3 × 3 = 54 = t₁ + t₂ + t₃…………(iii)

From(i), (ii) and (iii),

t₁ = 8, t₂ = 18, t₃ = 28

Common difference, d = 18 – 8 = 10

t_m = 168

⇒ a + (m – 1)d = 168

⇒ 8 + (m – 1)×10 = 168

⇒ (m – 1) × 10 = 160

⇒ m – 1 = 16

∴m = 17

t₂₀ = a + (20 – 1)d

= 8 + 19 × 10

= 8 + 190

= 198Question 43

The sum of first q terms of an A.P. is 63q – 3q². If its pth term -60, find the value of p. Also, find the 11^th term of this A.P.

Remark* – Question modified.Solution 43

Let the first term be ‘a’ and the common difference be ‘d’.

Sum of the first ‘q’ terms, S_q = 63q – 3q²

S₁ = 63 × 1 – 3 × 1² = 60 = t₁……..(i)

S₂ = 63 × 2 – 3 × 2² = 114 = t₁ + t₂…..(ii)

S₃ = 63 × 3 – 3 × 3² = 162 = t₁ + t₂ + t₃ …..(iii)

From (i), (ii) and (iii),

t₁ = 60

t₂ = 54

t₃ = 48

Common difference, d = 54 – 60 = – 6

t_p = a + (p – 1)d

⇒ -60 = 60 + (p – 1)( – 6)

⇒ – 120 = – 6(p – 1)

⇒ p – 1 = 20

∴p = 21

t₁₁ = 60 + (11 – 1)( – 6)

=60 + 10( – 6)

= 60 – 60

= 0

The 11^th term of the A.P. is 0.Question 44

The sum of first m terms of an A.P. is 4m² – m. If its nth term is 107, find the value of n. Also, find the 21^st term of this A.P.Solution 44

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

Sum of m terms of the A.P., S_m = 4m² – m

S₁ = 4 × 1² – 1 = 3 = t₁ …….(i)

S₂ = 4 × 2² – 2 = 14 = t₁ + t₂……..(ii)

S₃ = 4 × 3² – 3 = 33 = t₁ + t₂ + t_3­ …….(iii)

From (i), (ii) and (iii)

t₁ = 3, t₂ = 11, t₃ = 19

Common difference, d = 11 – 3 = 8

t_n = 107

⇒ a + (n – 1)d = 107

⇒ 3 + (n – 1)8 = 107

⇒ 8(n – 1) = 104

⇒ n – 1 = 13

∴n = 14

t₂₁ = 3 + (21 – 1)8 = 3 + 160 = 163Question 45

Solution 45

Question 46

If the sum of first n terms of an A.P. is then find its n^th term. Hence write its 20^th term.Solution 46

Question 47 (i)

Solution 47 (i)

Question 47 (ii)

If the sum of first n terms of an A.P. is n², then find its 10^th term.Solution 47 (ii)

Sum of first n terms of an AP is given by

As per the question, S_n = n²

Hence, the 10^th term of this A.P. is 19.Question 48

Solution 48

Question 49

Solution 49

Question 50 (i)

Solution 50 (i)

Question 50 (ii)

Solution 50 (ii)

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55 (i)

Solution 55 (i)

Question 55(ii)

Find the sum of all integers between 100 and 550 which are not divisible by 9.Solution 55(ii)

Question 55(iii)

Find the sum of all integers between 1 and 500 which are multiplies 2 as well as of 5.Solution 55(iii)

Question 55(iv)

Find the sum of all integers from 1 to 500 which are multiplies 2 as well as of 5.Solution 55(iv)

Question 55(v)

Find the sum of all integers from 1 to 500 which are multiples of 2 or 5.Solution 55(v)

Question 56 (i)

Solution 56 (i)

Question 56 (ii)

Solution 56 (ii)

Question 56 (iii)

Solution 56 (iii)

Question 56 (iv)

Solution 56 (iv)

Question 56 (v)

Solution 56 (v)

Question 56 (vi)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find

n and a_n, if a = 2, d = 8 and S_n = 90.Solution 56 (vi)

Question 56(vii)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find k, if S_n = 3n² + 5n and a_k = 164.Solution 56(vii)

Question 56 (viii)

Let there be an A.P. with first term ‘a’, common difference ‘d’. If a_n denotes its n^th term and S_n the sum of first n terms, find S₂₂, if d = 22 and a₂₂ = 149.Solution 56 (viii)

The nth term of an A.P. is given by a_n = a + (n – 1)d

Sum of first n terms of an AP is given by

Question 57

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

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 58

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 59

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 60

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 61

Trees planted by the student in class 1 = 2 + 2 = 4

Trees planted by the student in class 2 = 4 + 4 = 8

Trees planted by the students in class 3 = 6 + 6 = 12

…….

Trees planted by the students in class 12 = 24 + 24 = 48

∴ the series will be 4, 8, 12,………., 48

a = 4, Common Difference, d = 8 – 4 = 4

Let ‘n’ be the number of terms in the series.

48 = 4 + (n – 1)4

⇒ 44 = 4(n – 1)

⇒ n – 1 = 11

∴n = 12

Sum of the A.P. series,

Number of trees planted by the students = 312Question 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 62

Since, the difference between the savings of two consecutive months is Rs. 20, therefore the series is an A.P.

Here, the savings of the first month is Rs. 50

First term, a = 50, Common difference, d = 20

No. of terms = no. of months

No. of terms, n = 12

= 6[100 + 220]

= 6×320

= 1920

After a year, Ramakali will save Rs. 1920.

Yes, Ramakali will be able to fulfill her dream of sending her daughter to school.Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

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

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

R.H.S.

= 3(S₂₀ – S₁₀)

= 3(10[2a + 19d] – 5[2a + 9d])

= 3(20a + 190d – 10a – 45d)

= 3(10a + 145d)

= 3 × 5(2a + 29d)

= 15[2a + (30 – 1)d]

= S₃₀

= L.H.S.Question 71

Solve the equation

(-4) + (-1) + 2 + 5 + …. + x = 437.Solution 71

Question 72

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

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 73

Question 74

The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. 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 74