**Question **1. Which of the following sequences are in arithmetic progression?

(π) 2, 6, 10, 14,

(ππ) 15, 12, 9, 6,

(πππ) 5, 9, 12, 18,

(ππ£) 1/2, 1/3, 1/4, 1/5

Solution:

(π) 2, 6, 10, 14,

π = π_{2} β π1

π_{1} = 6 β 2

π_{1} = 4

π_{2} = 10 β 6

π_{2} = 4

π_{3} = 14 β 10

π_{3} = 4

Here,

π_{1} = π_{2} = π_{3}

Hence, the given sequence is in arithmetic progression.

(ππ) 15, 12, 9, 6,

π = π_{2} β π_{1}

π_{1} = 12 β 15

π_{1} = β3

π_{2} = 9 β 12

π_{2} = β3

π_{2} = 6 β 9

π_{3} = β3

Here,

π_{1} = π_{2} = π_{3}

Hence, the given sequence is in arithmetic progression.

(πππ) 5, 9, 12, 18,

π = π_{2} β π_{1}

π_{1} = 9 β 5

π_{1} = 4

π_{2} = 12 β 9

π_{2} = 3

π_{1} β π_{2}

Hence, the given sequence is not in arithmetic progression

**Question **2. The ππ‘β term of sequence is (2πβ 3), find its fifteenth term.

Solution:

It is given that,

ππ‘β term of sequence is (2πβ 3)

So, 15^{π‘β} term is

2(15)β 3

30 β 3 = 27

Hence, the fifteenth term is 27.

**Question **3. If the ππ‘β term of an A.P. is (2π + 3), find the A.P.

Solution:

It is given that,

p^{π‘β} term of sequence is (2π + 3)

So, 1^{π π‘} term is

2(1) + 3

2 + 3 =5

So, 1^{π π‘} term is

2(1) + 3

2 + 3 =5

So, 2^{ππ} term is

2(2) + 3

4 + 3 =7

So, 3^{ππ }term is

2(3) + 3

6 + 3 = 9

Hence, the AP is 5, 7, 9,β¦β¦β¦β¦ .

**Question **4. Find the 24^{π‘β} term of the sequence: 12, 10, 8, 6, β¦ β¦ β¦

Solution:

It is given that,

12, 10, 8, 6, β¦ β¦ β¦ β¦ β¦ ..

π = 10 β 12

π = β2

π = 12

π = 24

π^{π‘β} = π + (π β 1)π

24^{π‘β} = 12 + (24 β 1)(β2)

24^{π‘β} = 12 + (23)(β2)

24^{π‘β} = 12 + (23)(β2)

24^{π‘β} = 12 β 46

24^{π‘β} = β34

Hence, the 24^{π‘β} term is β34.

**Question **5. Find the 30^{π‘β} term of the sequence:

**Question **6. Find the 100^{π‘β} term of the sequence β3, 2β3, 3β3, β¦ β¦ β¦ β¦ β¦ . .. .

Solution:

It is given that,

β3, 2β3, 3β3, β¦ β¦ β¦ β¦ β¦

π = 2β3 β β3

π = β3

π = β3

π = 100^{π‘β}

π^{π‘β} = π + (π β 1)π

100^{π‘β} = β3 + (100 β 1)β3

100^{π‘β} = β3 + 100β3 β β3

100^{π‘β} = 100β3

Hence, the 100^{π‘β} term of the given AP is 100β3.

**Question **7. Find the 50^{π‘β }term of the sequence:

**Question **8. Is 402 a term of the sequence: 8, 13, 18, 23,β¦β¦β¦β¦.?

Solution:

It is given that,

8, 13, 18, 23, β¦β¦β¦β¦.

π = 13 β 8

π = 5

π = 8

π^{π‘β} = 402

π^{π‘β} = π + (π β 1)π

402 = 8 + (π β 1)5

402 = 8 + 5π β 5

402 = 3 + 5π

402 β 3 = 5π

399 = 5π

π = 399/5

Hence, 402 is not a term of the given sequence.

**Question **9. Find the common difference and 99π‘β term of the arithmetic progression:

**Question **10. How many terms are there in the series:

(i) 4, 7, 10, 13, β¦β¦β¦β¦ 148

(ii) 0.5, 0.53, 0.56, β¦β¦β¦β¦.1.1

(iii) 3/4 , 1,1 (1/4 ), β¦ β¦ β¦ β¦ . ,3

Solution:

(i) It is given that,

4, 7, 10, 13, β¦β¦β¦β¦ 148

π = 7 β 4

π = 3

π = 4

π^{π‘β} = 148

π^{π‘β }= π + (π β 1)π

148 = 4 + (π β 1)3

148 = 4 + 3π β 3

148 = 1 + 3π

148 β 1 = 3π

147 = 3π

π = 147/3

π = 49

Hence, there are total 49 terms in the given AP.

(ii) It is given that,

0.5, 0.53, 0.56, β¦β¦β¦β¦.1.1

π = 0.53 β 0.50

π = 0.03

π = 0.5

π^{π‘β} = 1.1

π^{π‘β} = π + (π β 1)π

1.1 = 0.5 + (π β 1)0.03

1.1 = 0.5 + 0.03π β 0.03

1.1 β 0.5 = 0.03π β 0.03

0.6 = 3π β 0.03

0.6 + 0.03 = 0.03π

0.63 = 0.03π

π = 0.63/0.03

π = 21

Hence, there are total 21 terms in the given AP.

**Question **11. Which term of the A.P. 1 + 4 + 7 + 10 +β¦β¦β¦β¦ is 52?

Solution:

It is given that,

1 + 4 + 7 + 10 +β¦β¦β¦β¦

π = 4 β 1

π = 3

π = 1

π^{π‘β} = 52

π^{π‘β} = π + (π β 1)π

52 = 1 + (π β 1)3

52 β 1 = (π β 1)3

51 = (π β 1)3

51/3 = π β 1

17 = π β 1

17 + 1 = π

π = 18

Hence, there are total 18 terms in the given AP.

**Question **12. If 5π‘β and 6^{π‘β} terms of an A.P are respectively 6 and 5. Find the 11^{π‘β} term of the A.P.

Solution:

We know that,

π^{π‘β} = π + (π β 1)π

5^{π‘β} = 6

π + (5 β 1)π = 6

π + 4π = 6_____________(π)

6^{π‘β} = 5

π + (6 β 1)π = 5

π + 5π = 5_____________(ππ)

Subtracting (ii) from (i), we get

(π + 5π) β (π + 4π) = 5 β 6

π + 5π β π β 4π = 5 β 6

5π β 4π = 5 β 6

π = β1

Put the value of d in equation (i)

π + 4(β1) = 6

π β 4 = 6

π = 6 + 4

π = 10

Now,

π = 11

π^{π‘β} = 10 + (11 β 1)(β1)

π^{π‘β} = 10 + 10(β1)

π^{π‘β} = 10 β 10

π^{π‘β} = 0

Hence, the 11th term of the AP is 0.

**Question **13. If π‘π represents π^{π‘β} term of an A.P, π‘_{2} + π‘_{5} β π‘_{3} = 10 and π‘_{2} + π‘_{9} = 17, find its first term and its common difference

Solution:

We know that,

π^{π‘β }= π + (π β 1)π

π‘_{2} = π + (2 β 1)π

π‘_{2} = π + π

π‘_{5 }= π + (5 β 1)π

π‘_{5} = π + 4π

π‘_{3} = π + (3 β 1)π

π‘_{3} = π + 2π

According to question,

π‘_{2} + π‘_{5} β π‘_{3} = 10

(π + π) + (π + 4π) β (π + 2π) = 10

π + π + π + 4π β π β 2π = 10

π + 3π = 10_______________(π)

Again,

π‘_{2} = π + (2 β 1)π

π‘_{2} = π + π

π‘_{2} = π + (9 β 1)π

π‘_{2} = π + 8π

According to question,

π‘_{2} + π‘_{9} = 17

π + π + π + 8π = 17

2π + 9π = 17______________(ππ)

From equation (i) we get,

π + 3π = 10

π = 10 β 3π** __**(πππ)

Put the value of a in equation (ii)

2(10 β 3π) + 9π = 17

20 β 6π + 9π = 17

20 + 3π = 17

20 + 3π = 17

3π = 17 β 20

3π = β3

π = β1

Put the value of d in equation (iii)

π = 10 β 3(β1)

π = 10 + 3

π = 13

Hence, the value of π is 13 and value of π is β1.

**Question **14. Find the 10^{π‘β} term from the end of the A.P. 4, 9, 14,β¦β¦β¦β¦β¦..,254.

Solution:

It is given that,

4, 9, 14, β¦ β¦ β¦ β¦ β¦ . . , 254

π = 4

π = 9 β 4

π = 5

π = 254

Now,

10^{th} term from the end,

π_{10} = π β (π β 1)π

π_{10} = 254 β (10 β 1)5

π_{10} = 254 β (9)5

π_{10} = 254 β 45

π_{10} = 209

Hence, the 10π‘β term is 209.

**Question **15. Determine the arithmetic progression whose 3^{ππ} term is 5 and 7^{π‘β} term is 9.

Solution:

It is given that,

3ππ term = 5

7π‘β term = 9

From 3^{ππ} term we get,

3^{ππ} term = 5

π + (3 β 1)π = 5

π + 2π = 5_______________(π)

From 7th term we get,

π + (7 β 1)π = 5

π + 6π = 9______________(ππ)

From equation (i) we get,

π + 2π = 5

π = 5 β 2π** __**(πππ)

Put the value of π in equation (ii)

5 β 2π + 6π = 9

5 + 4π = 9

4π = 9 β 5

4π = 4

π = 4/4

π = 1

Put the value of d in equation (iii)

π = 5 β 2(1)

π = 5 β 2

π = 3

π

_{1}= 3

π

_{2}= 3 + 1 = 4

π

_{3}= 3 + 2(1) = 5

π

_{4}= 3 + 3(1) = 6

Hence, the required AP is 3,4,5,6, β¦ β¦ β¦ β¦

**Question **16. Find the 31^{π π‘} term of an A.P whose 10^{π‘β} term is 38 and 16^{π‘β} term is 74.

Solution:

It is given that,

10^{ππ} term = 38

16^{π‘β} term = 74

From 10^{th} term we get,

10^{π‘β} term = 38

π + (10 β 1)π = 38

π + 9π = 38_______________(π)

From 16th term we get,

π + (16 β 1)π = 74

π + 15π = 74______________(ππ)

From equation (i) we get,

π + 9π = 38

π = 38 β 9π** __**(πππ)

Put the value of π in equation (ii)

38 β 9π + 15π = 74

38 + 6π = 74

6π = 74 β 38

6π = 36

π = 36/6

π = 6

Put the value of d in equation (iii)

π = 38 β 9(6)

π = 38 β 54

π = β16

To find 31st term,

π

^{π‘β}= π + (π β 1)π

31

^{π‘β}= β16 + (31 β 1)6

31

^{π‘β}= β16 + (30)6

31

^{π‘β}= β16 + 180

31

^{π‘β}= 164

Hence, the 31

^{st}term is 164.

**Question **17. Which term of the services?

21, 18, 15, β¦β¦β¦β¦β¦β¦β¦β¦β¦ is β 81?

Can any term of this series be zero? If yes find the number of term.

Solution:

It is given that,

21, 18, 15,β¦β¦β¦β¦β¦β¦β¦β¦.β 81

π = 18 β 21

π = β3

π = 21

π^{π‘β} = β81

π^{π‘β} = π + (π β 1)π

β81 = 21 + (π β 1)(β3)

β81 β 21 = β3π + 3

β102 = β3π + 3

β102 β 3 = β3π

β105 = β3π

3π = 105

π = 105/3

π = 35

Hence, the 35^{th }term of the AP is -81.

Now,

π^{π‘β} = π + (π β 1)π

0 = 21 + (π β 1)(β3)

0 β 21 = β3π + 3

β21 = β3π + 3

β21 β 3 = β3π

β24 = β3π

3π = 24

π = 24/3

π = 8

Hence, the 8th term of the AP is 0.

**Question **18. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31^{st} term.

Solution:

It is given that,

π^{π‘β} = 60

π = 7

π = 125

π^{π‘β} = π + (60 β 1)π

125 = 7 + 59π

118 = 59π

59π = 118

π = 2

Now,

π^{π‘β} = π + (π β 1)π

π^{π‘β} = 7 + (31 β 1)(2)

π^{π‘β} = 7 + 30 Γ 2

π^{π‘β} = 7 + 60

π^{π‘β} = 67

Hence, the value of 31st term is 67.

**Question **19. The sum of the 4^{th} and the 8^{th} terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34.

Find the first three terms of the A.P.

Solution:

It is given that,

π_{4} + π8 = 24

π + (4 β 1)π + π + (8 β 1)π = 24

(π + 3π) + (π + 7π) = 24

2π + 10π = 24

π + 5π = 12 β¦ β¦ β¦ β¦ β¦ . (π)

And,

π_{6} + π10 = 34

π + (6 β 1)π + π + (10 β 1)π = 24

(π + 5π) + (π + 9π) = 34

2π + 14π = 34

π + 7π = 17 β¦ β¦ β¦ β¦ . (ππ)

From equation (i) we get,

π = 12 β 5π

Put the value of a in equation (ii)

12 β 5π + 7π = 17

β5π + 7π = 17 β 12

2π = 5

π = 5/2

Put the value of βdβ in (iii)

**Question **20. If the third term of an A.P. is 5 and the seventh terms are 9, find the 17^{th} term.

Solution:

It is given that,

π_{2} = 5

π + 2π = 5 β¦ β¦ β¦ . β¦ . (π)

π_{7} = 9

π + 6π = 9 β¦ β¦ β¦ β¦ β¦ β¦ (ππ)

From equation (i), we get

π + 2π = 5

π = 5 β 2π

Put the value a in equation (ii)

5 β 2π + 6π = 9

β2π + 6π = 9 β 5

4π = 4

π = 1

Put the value of d in equation (i)

π + 2(1) = 5

π = 3

Now,

17^{π‘β} = π + (π β 1)π

17^{π‘β} = 3 + (17 β 1)(1)

17^{π‘β} = 3 + 16 Γ 1

17^{π‘β} = 19

Hence, the value of 31^{st} term is 67.

### Exercise 10B

**Question **1. In an A.P., ten times of its tenth term is equal to thirty times of its 30^{th} term. Find its 40^{th} term.

Solution:

It is given that,

10 Γ π_{10} = 30 Γ π_{30}

10 Γ π_{10 }= 30 Γ π_{30}

10 Γ [π + (10 β 1)π] = 30 Γ [π + (30 β 1)π]

10 Γ [π + 9π] = 30 Γ [π + 29π]

[π + 9π] = 3[π + 29π]

π + 9π = 3π + 87π

π β 3π = 87π β 9π

β2π = 78π

π = β78π/2

π = β39π

To find π_{40}

π_{40} = β39π + (40 β 1)π

π_{40} = β39π + 39π

π_{40} = 0

Hence, the value of π_{40} is 0.

**Question **2. How many two-digit numbers are divisible by 3?

Solution:

It is given that,

Two-digit numbers are divisible by 3, so the required AP.

12,15,18,21, β¦ β¦ β¦ β¦ β¦ β¦ β¦ 99

π = 12

π = 15 β 12

π = 3

π^{π‘β} term = 99

π^{π‘β} = π + (π β 1)π

99 = 12 + (π β 1)3

99 β 12 = (π β 1)3

87 = (π β 1)3

87/3 = π β 1

29 = π β 1

29 + 1 = π

30 = π

π = 30

Hence, there are 30 two digit number divisible by 3.

**Question **3. Which term of A.P. 5, 15, 25β¦β¦β¦β¦β¦β¦..will be 130 more than its 31^{st }term?

Solution:

It is given that,

5, 15, 25 β¦ β¦ β¦ β¦ β¦ β¦ ..

π = 5

π = 15 β 5

π = 10

π = 31

π^{π‘β} = π + (π β 1)π

π_{31} = 5 + (31 β 1)10

π_{31} = 5 + (30)10

π_{31} = 5 + 300

π_{31} = 305/Now,

π^{π‘β} βπ31 = 130

5 + (π β 1)10 β 305 = 130

(π β 1)10 β 300 = 130

(π β 1)10 = 130 + 300

(π β 1)10 = 430

π β 1 = 430/10

π β 1 = 43

π = 43 + 1

π = 44

**Question **4. Find the value of p, if x, 2x + p and 3x + 6 are in A.P.

Solution:

It is given that,

π₯, 2π₯ + π and 3π₯ + 6 are in A.P.

(2π₯ + π) β π₯ = (3π₯ + 6) β (2π₯ + π)

2π₯ + π β π₯ = 3π₯ + 6 β 2π₯ + π

π₯ + π = π₯ + 6 β π

π = 6 β π

π + π = 6

2π = 6

π = 6/2

π = 3

Hence, the value of p is 3.

**Question **5. If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, which term of it, is zero?

Solution:

It is given that,

3^{ππ }π‘πππ = 4

π + (3 β 1)π = 4

π + 2π = 4 β¦ β¦ β¦ β¦ β¦ (π)

9^{π‘β} π‘πππ = β8

π + (9 β 1)π = β8

π + 8π = β8 β¦ β¦ β¦ β¦ β¦ (ππ)

From equation (i) we get,

π = 4 β 2π β¦ β¦ β¦ β¦ β¦ (πππ)

Put the value of a in equation (ii)

4 β 2π + 8π = β8

6π = β8 β 4

6π = β12

π = β12/6

π = β2

Put the value of d in equation (iii)

π = 4 β 2(β2)

π = 4 + 4

π = 8

Let us assume that,

π^{π‘β} = π + (π β 1)π

0 = 8 + (π β 1)(β2)

β8 = β2π + 2

β8 β 2 = β2π

β10 = β2π

β10/β2 = π

π = 5

Hence, 5^{th} term of the given AP is 0.

**Question **6. How many three-digit numbers are divisible by 87?

Solution:

It is given that,

Three-digit numbers are divisible by 87

174, 261, β¦ β¦ β¦ β¦ β¦ β¦ . , 957

π = 174

π = 261 β 174

π = 87

π^{π‘β} π‘πππ = 957

π^{π‘β} = π + (π β 1)π

957 = 174 + (π β 1)87

957 β 174 = (π β 1)87

783 = (π β 1)87

783/87 = π β 1

9 = π β 1

9 + 1 = π

π = 10

Hence, there are 10 three-digit numbers are divisible by 87.

**Question **7. For what value of n, the nth term of A.P 63, 65, 67,β¦β¦β¦β¦. and nth term of A.P. 3, 10, 17,β¦.β¦β¦are equal to each other?

Solution:

It is given that,

63, 65, 67, β¦ β¦ β¦ β¦ .

π = 63

π = 65 β 63

π = 2

π^{π‘β} = π + (π β 1)π

π^{π‘β} = 63 + (π β 1)2

Again,

3,10,17, β¦ . β¦ β¦

π = 3

π = 10 β 3

π = 7

π^{π‘β} = π + (π β 1)π

π^{π‘β} = 3 + (π β 1)7

By comparing both the terms,

63 + (π β 1)2 = 3 + (π β 1)7

63 + 2π β 2 = 3 + 7π β 7

61 + 2π = 7π β 4

2π β 7π = β4 β 61

β5π = β65

π = β65/β5

π = 13

Hence, the value of n is 13.

**Question **8. Determine the A.P. whose 3^{rd} term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.

Solution:

It is given that,

3^{ππ }π‘πππ = 16

π + (3 β 1)π = 16

π + 2π = 16_______________(π)

Also,

7^{π‘β} π‘πππ β 5^{π‘β} π‘πππ = 12

π + (7 β 1)π β [π + (5 β 1)π] = 12

π + 6π β π β 4π = 12

6π β 4π = 12

2π = 12

π = 6

Put the value of d in (i) equation,

π + 2(6) = 16

π + 12 = 16

π = 16 β 12

π = 4

First term of AP is 4

Second term of AP is π + π = 4 + 6 = 10

Third term of AP is π + 2π = 4 + 2(6) = 4 + 12 = 16

Hence, the AP is 4, 10, 16 β¦ β¦ β¦ β¦ ..

**Question **9. If numbers πβ 2,4πβ 1 and 5π + 2 are in A.P. find the value of n and its next two terms.

Solution:

It is given that,

πβ 2, 4πβ 1 and 5π + 2 are in A.P.

(4πβ 1) β (πβ 2) = (5π + 2) β (4πβ 1)

4πβ 1 β π + 2 = 5π + 2 β 4π + 1

3π + 1 = π + 3

3π β π = 3 β 1

2π = 2

π = 1

First term = 1β 2

= β1

Second term = 4πβ 1

= 4(1) β 1

= 4 β 1

= 3

Third term = 5π + 2

= 5(1) + 2

= 5 + 2

= 7

So the AP is -1, 3, 7β¦β¦β¦.

4^{π‘β} π‘πππ = π + (π β 1)π

= β1 + (4 β 1)4

= β1 + (3)4

= β1 + 12

= 11

5^{π‘β} π‘πππ = π + (π β 1)π

= β1 + (5 β 1)4

= β1 + (4)4

= β1 + 16

= 15

Hence, the next terms is 11 and 15

**Question **10. Determine the value of k for which π^{2} + 4π + 8, 2π^{2} + 3π + 6 and 3π^{2} + 4π + 4 are in A.P.

Solution:

It is given that,

π^{2} + 4π + 8, 2π^{2} + 3π + 6 and 4π^{2} + 4π + 4 are in A.P.

(2π^{2} + 3π + 6) β (π^{2} + 4π + 8) = (3π^{2} + 4π + 4) β (2π^{2} + 3π + 6)

2π^{2} + 3π + 6 β π^{2} β 4π β 8 = 3π^{2} + 4π + 4 β 2π^{2} β 3π β 6

π^{2} β π β 2 = π^{2} + π β 2

βπ β π = β2 + 2

β2π = 0

π = 0

Hence, the value of k is 0.

**Question **11. If a, b and c are in A.P show that:

(i) 4a, 4b and 4c are in A.P

(ii) a + 4, b + 4 and c + 4 are in A.P.

Solution:

It is given that,

a, b and c are in A.P.

π β π = π β π

π + π = π + π

2π = π + π

π = π+π/2

(i) 4a, 4b and 4c are in A.P

By common difference,

4π β 4π = 4π β 4π

2(2π β 2π) = 2(2π β 2π)

2(π + π β 2π) = 2(2π β (π + π))

2(π β π) = 2(2π β π β π)

2(π β π) = 2(π β π)

Hence, the given term is in A.P.

(ii) a + 4, b + 4 and c + 4 are in A.P.

By common difference,

(π + 4) β (π + 4) = (π + 4) β (π + 4)

(π + 4) β (π + 4) = (π + 4) β (π + 4)

π + 4 β π β 4 = π + 4 β π β 4

π β π = π β π

Put the value of b,

**Question **12. An A.P consists of 57 terms of which 7^{π‘β} term is 13 and the last term is 108. Find the 45^{th} term of this A.P.

Solution:

It is given that,

Total number of terms is 57.

7^{π‘β }π‘πππ = 13

π + (7 β 1)π = 13

π + 6π = 13_______________(π)

57^{π‘β} π‘πππ = 108

π + (57 β 1)π = 108

π + 56π = 108_______________(ππ)

From equation (i) we get,

π = 13 β 6π** _**(πππ)

Put the value of a in equation (ii)

(13 β 6π) + 56π = 108

13 β 6π + 56π = 108

50π = 108 β 13

50π = 95

**Question **13. 4^{th} term of an A.P is equal to 3 times its first term and 7^{th} term exceeds twice the 3^{rd} time by 1. Find the first term and the common difference.

Solution:

It is given that,

4^{π‘β} π‘πππ = 3 Γ 1^{π π‘} π‘πππ

π + (4 β 1)π = 3π

π + 3π = 3π

π β 3π + 3π = 0

2π β 3π = 0________________(π)

Again,

7^{π‘β} π‘πππ β 2(3^{ππ} π‘πππ) = 1

π + (7 β 1)π β 2(π + (3 β 1)π) = 1

π + 6π β 2(π + 2π) = 1

π + 6π β 2π β 4π = 1

βπ + 2π = 1________________(ππ)

From equation (i) get the value of a,

2π = 3π

**Question **14. The sum of the 2^{nd} term and the 7^{th} term of an A.P is 30. If its 15^{th} term is 1 less than twice of its 8th term, find the A.P.

Solution:

It is given that,

2^{π‘β} π‘πππ + 7^{π π‘} π‘πππ = 30

π + (2 β 1)π + π + (7 β 1)π = 30

π + π + π + 6π = 30

2π + 7π = 30______________(π)

Again,

2 Γ 8^{π‘β} π‘πππ β 15^{π π‘} π‘πππ = 1

2 Γ [π + (8 β 1)π] β [π + (15 β 1)π] = 1

2 Γ [π + 7π] β (π + 14π) = 1

2π + 14π β π β 14π = 1

π = 1

Put the value of a in equation (i)

2π + 7π = 30

2(1) + 7π = 30

2 + 7π = 30

7π = 30 β 2

7π = 28

π = 28/7

π = 4

First term = 1

Second term = 1 + 4 = 5

Third term = 1 + 2(4) = 1 + 8 = 9

Hence, the required AP is 1, 5, 9,β¦β¦.. .

**Question **15. In an A.P, if m^{th }term is n and n^{th} term is m, show that its r^{th} term is (m + n β r)

Solution:

It is given that,

π^{π‘β} π‘πππ = π

π + (π β 1)π = π(π)

π^{π‘β} π‘πππ = π

π + (π β 1)π = π(ππ)

By subtracting (i) equation from (ii) equation, we get

π + (π β 1)π β [π + (π β 1)π] = π β π

π + (π β 1)π β π β (π β 1)π = π β π

(π β 1)π β (π β 1)π = π β π

π_{π} β π β ππ + π = π β π

π_{π} β ππ = π β π

π(π β π) = π β π

βπ(π β π) = π β π

βπ = πβπ/πβπ

π = β1

Put the value of d in equation (i)

π + (π β 1)(β1) = π

π β π + 1 = π

π = π + π β 1

According to question,

π^{π‘β} π‘πππ

π + (π β 1)π

(π + π β 1) + (π β 1)(β1)

π + π β 1 β π + 1

π + π β π

Hence proved.

**Question **16. Which term of the A.P 3, 10, 17,β¦β¦β¦β¦ will be 84 more than its 13^{th} term?

Solution:

It is given that,

3, 10, 17, β¦ β¦ β¦ β¦

π = 3

π = 10 β 3

π = 7

To find 13^{th} term,

13^{π‘β} π‘πππ = 3 + (13 β 1)7

13^{π‘β} π‘πππ = 3 + (12)7

13^{π‘β} π‘πππ = 3 + 84

13^{π‘β} π‘πππ = 87

According to question,

π^{π‘β} π‘πππ β 13^{π‘β} π‘πππ = 84

[3 + (π β 1)7] β 87 = 84

3 + (π β 1)7 = 84 + 87

3 + (π β 1)7 = 171

(π β 1)7 = 171 β 3

(π β 1)7 = 168

π β 1 = 168/7

π β 1 = 24

π = 24 + 1

π = 25

Hence, the required term is 25.

### Exercise 10 C

**Question **1. Find the sum of the first 22 terms of the A.P.: 8, 3, β2, β¦ β¦ β¦ ..

Solution:

It is given that,

8, 3, β2, β¦ β¦ β¦ . .

π = 8

π = 3 β 8

π = β5

π = 22

π_{π} = π/2 [2π + (π β 1)π]

π_{22} = 22/2[2(8) + (22 β 1)(β5)]

π_{22} = 11[16 + 21(β5)]

π_{22} = 11[16 β 105]

π_{22} = 11 Γ (β89)

π_{22} = β979

Hence, the sum of first 22 term is -979.

**Question **2. How many terms of the A.P.: 24, 21, 18,β¦β¦β¦β¦. must be taken so that their sum is 78?

Solution:

It is given that,

24, 21, 18, β¦ β¦ β¦ β¦.

π = 24

π = 21 β 24

π = β3

π_{π} = 78

π_{π} = π/2 [2π + (π β 1)π]

78 = π/2 [2(24) + (π β 1)(β3)]

78 Γ 2 = π[48 + (π β 1)(β3)]

156 = π[48 + (π β 1)(β3)]

156 = π[48 + (β3π + 3)]

156 = π[48 β 3π + 3]

156 = π[51 β 3π]

156 = 51π β 3π^{2}

51π β 3π^{2} β 156 = 0

β3(β17π + π^{2} + 52) = 0

π^{2} β 17π + 52 = 0

π^{2} β 13π β 4π + 52 = 0

π(π β 13) β 4(π β 13) = 0

(π β 4)(π β 13) = 0

π β 4 = 0 π β 13 = 0

π = 4 π = 13

**Question **3. Find the sum of 28 terms of an A.P. whose nth term is 8πβ 5.

Solution:

It is given that,

nth term of an AP is 8πβ 5

First term = 8(1)β 5

= 8β 5

= 3

Second term = 8(2)β 5

= 16β 5

= 11

π = 3

π = 11 β 3

π = 8

π^{π‘β} = 28

π_{π} = π/2 [2π + (π β 1)π]

π_{π} = 28/2 [2(3) + (28 β 1)8]

π_{π} = 14[6 + (27)8]

π_{π} = 14[6 + 216]

π_{π} = 14[222]

π_{π} = 3108

Hence, the required sum of 28 terms is 3108.

**Question **4 (i). Find the sum of all odd natural numbers less than 50.

Solution:

It is given that,

1, 3, 5, 7, 9 β¦ β¦ β¦ β¦ β¦ β¦ .49

π = 3

π = 3 β 1

π = 2

π^{π‘β} = π + (π β 1)π

49 = 1 + (π β 1)2

49 β 1 = (π β 1)2

48 = (π β 1)2

48/2 = π β 1

24 = π β 1

π = 24 + 1

π = 25

Sum of all odd numbers less than 50,

**Question **4 (ii). Find the sum of first 12 natural numbers each of which is a multiple of 7.

Solution:

According to the question,

Natural numbers multiple of 7 are,

7, 14, 21, 28, 35 β¦ β¦ β¦ β¦ β¦ β¦ β¦ β¦ . .84

Here,

π = 7

π = 84

Sum of 12 natural numbers multiple of 7 are,

**Question **5. Find the sum of first 51 terms of an A.P. whose 2^{nd} and 3^{rd} terms are 14 and 18 respectively.

Solution:

πΌπ‘ ππ πππ£ππ π‘βππ‘,

2^{ππ }π‘πππ = 14

π + (2 β 1)π = 14

π + π = 14_________________(π)

3^{ππ} π‘πππ = 18

We know that,

π = 18 β 14

π = 4

Put the value of d in equation (i)

π + π = 14

π + 4 = 14

π = 14 β 4

π = 10

To find 51 term,

**Question **6. The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Solution:

It is given that,

Sum of first 7 terms = 49

π_{7} = 49

**Question **7. The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.

Solution:

It is given that,

First term of an AP is 5

Last term of an AP is 45

Sum of terms is 1000

**Question **8. Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Solution:

We know that,

Natural numbers between 250 and 1000 are,

252, 261, 270, β¦ β¦ β¦ β¦ .999

Total terms in AP are,

π = π + (π β 1)π

999 = 252 + (π β 1)9

999 β 252 = (π β 1)9

747 = (π β 1)9

747/9 = (π β 1)

83 = (π β 1)

83 + 1 = π

π = 84

Sum of natural numbers between 250 and 1000.

**Question **9. The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Solution:

It is given that,

π = 34

π = 18

π = 700

π = π + (π β 1)π

700 = 34 + (π β 1)18

700 β 34 = (π β 1)18

666 = (π β 1)18

666/18 = π β 1

37 = π β 1

37 + 1 = π

38 = π

π = 38

Sum of first 38 terms,

π_{π} = π/2 [π + π]

π_{38} = 38/2 [34 + 700]

π_{38} = 19 Γ 734

π_{38} = 13946

Hence, the sum of first 38 terms is 13946.

**Question **10. In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.

Solution:

It is given that,

π = 25

π = β17

Sum of n terms = 132

**Question **11. If the 8^{th }term of an A.P is 37 and the 15^{th} term is 15 more than the 12^{th} term, find the A.P. Also, find the sum of first 20 terms of A.P.

Solution:

It is given that,

8^{π‘β} π‘πππ = 37

π + (8 β 1)π = 37

π + 7π = 37_____________(π)

15^{π‘β} π‘πππ β 12^{π‘β} π‘πππ = 15

π + (15 β 1)π β π + (12 β 1)π = 15

π + 14π β π + 11π = 15

14π β 11π = 15

3π = 15

π = 15/3

π = 5

Put the value of d in equation (i)

π + 7(5) = 37

π + 35 = 37

π = 37 β 35

π = 37 β 35

π = 2

First term = 2

Second term = 2 + 5 = 7

Second term = 2 + 2(5)

= 2 + 10

= 12

So, the required AP is 2,7,12 β¦ β¦ ..

Sum of first 20 terms,

**Question **12. Find the sum of all multiples of 7 between 300 and 700.

Solution:

We know that,

Numbers between 300 and 700 multiple of 7 are,

301,308,315 β¦ β¦ β¦ β¦ .693

Total terms in AP are,

π = 301

π = 308 β 301

π = 7

π_{π} = π + (π β 1)π

693 = 301 + (π β 1)7

693 β 301 = (π β 1)7

392 = (π β 1)7

392/7 = (π β 1)

56 = (π β 1)

56 + 1 = π

π = 57

Sum of first 57 terms,

**Question **13. The sum of n natural numbers is 5π^{2} + 4π. Find its 8^{th} term.

Solution:

It is given that,

Sum of n natural numbers is 5π^{2} + 4π

ππ = 5π^{2} + 4π

π_{πβ1} = 5(π β 1)^{2} + 4(π β 1)

= 5(π^{2} + 1 β 2π) + 4π β 4

= 5π^{2} + 5 β 10π + 4π β 4

= 5π^{2} β 6π + 1

π^{π‘β} π‘πππ = ππ β ππβ1

= 5π^{2} + 4π β (5π^{2} β 6π + 1)

= 5π^{2} + 4π β 5π^{2} + 6π β 1

= 10π β 1

To find 8^{th} term,

10π β 1

10(8) β 1

80 β 1 = 79

Hence, the 8th term is 79.

**Question **14. The fourth term of an A.P. is 11 and the term exceeds twice the fourth term by 5 the A.P and the sum of first 50 terms.

Solution:

It is given that,

4^{π‘β} π‘πππ = 11

π + (4 β 1)π = 11

π + 3π = 11____________(π)

Also,

8^{π‘β }π‘πππ β 2(4^{π‘β} π‘πππ) = 5

π + (8 β 1)π β 2(11) = 5

π + 7π β 22 = 5

π + 7π = 5 + 22

π + 7π = 27____________(ππ)

From equation (i) we get,

π + 3π = 11

π = 11 β 3π**_**(πππ)

Put the value of a in equation (ii)

11 β 3π + 7π = 27

4π = 27 β 11

4π = 16

π = 16/4

π = 4

Put the value d in equation (iii)

π = 11 β 3(4)

π = 11 β 12

π = β1

First term = β1

Second term = β1 + 4 = 3

Third term = β1 + 2(4) = β1 + 8 = 7

So, the required AP is β1,3,7, β¦ β¦ β¦ β¦ β¦ β¦.

Sum of 50 terms,

π

_{π }= π/2 [2π + (π β 1)π]

π

_{50}= 50/2 [2(β1) + (50 β 1)4]

π

_{50}= 25[β2 + (49)4]

π

_{50}= 25[β2 + 196]

π

_{50}= 25[194]

π

_{50}= 4850

Hence, the sum of 50 terms is 4850.

### Exercise 10 D

**Question **1. Find three numbers in A.P. whose sum is 24 and whose product is 440.

Solution:

Let us assumed that,

Three number in AP is π β π, π and π + π

According to the question,

π β π + π + π + π = 24

3π = 24

π = 24/3

π = 8

And,

(π β π) Γ π Γ (π + π) = 440

(π^{2} β π^{2}) Γ π = 440

Put the value of a,

(82 β π^{2}) Γ 8 = 440

(64 β π^{2}) Γ 8 = 440

(64 β π^{2}) = 440/8

(64 β π^{2}) = 55

π^{2} = 55 β 64

π^{2} = 9

π = β9

π = Β±3

Now,

If π = 3

π β π, π and π + π

8 β 3, 8 and 8 + 3

5, 8 and 11

If π = β3

8 β (β3),8 and 8 + (β3)

8 + 3, 8 and 8 β 3

11, 8 and 5.

**Question **2. The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.

Solution:

Let us assumed that,

Three number in AP is π β π, π and π + π

According to the question,

π β π + π + π + π = 24

3π = 21

π = 21/3

π = 7

And,

(π β π)^{2} + π^{2} + (π + π)^{2} = 165

π^{2} β 2ππ + π^{2} + π^{2} + π^{2} + 2ππ + π^{2} = 165

π^{2} + π^{2} + π^{2} + π^{2} + π^{2 }= 165

3π^{2} + 2π^{2} = 165

Put the value of a,

3π^{2 }+ 2π^{2} = 165

3(7)^{2} + 2π^{2} = 165

3 Γ 49 + 2π^{2} = 165

147 + 2π^{2} = 165

2π^{2} = 165 β 147

2π^{2} = 18

π^{2} = 18/2

π^{2} = 9

π = β9

π = Β±3

Now,

If π = 3

π β π, π and π + π

7 β 3, 7 and 7 + 3

4, 7 and 10

If π = β3

7 β (β3),9 and 7 + (β3)

7 + 3, 9 and 7 β 3

10, 7 and 4.

**Question **3. The angles of a quadrilateral are in A.P. with common difference 20Β°. Find its angles.

Solution:

Let us assumed that,

Four angles of a quadrilateral in A.P. π, π + 20Β°, π + 40Β° πππ π + 60Β°

By the angle sum property we know that,

π + (π + 20Β°) + (π + 40Β°) + (π + 60Β°) = 360Β°

π + π + 20Β° + π + 40Β° + π + 60Β° = 360Β°

4π + 20Β° + 40Β° + 60Β° = 360Β°

4π + 120Β° = 360Β°

4π = 360Β° β 120Β°

4π = 240Β°

π = 240Β°/4

π = 60Β°

First angle = 60Β°

Second angle = a + 20Β° = 60Β° + 20Β° = 80Β°

Third angle = a + 40Β° = 60Β° + 40Β° = 100Β°

Fourth angle = a + 40Β° = 60Β° + 60Β° = 120Β°

**Question **4. Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.

Solution:

Let us assumed that four parts be,

(π β 3π), (π β π), (π + π) πππ (π + 3π)

(π β 3π) + (π β π) + (π + π) + (π + 3π) = 96

π β 3π + π β π + π + π + π + 3π = 96

4π = 96

π = 96/4

π = 24

π^{2} = 36

π = β36

π = Β±6

If π = 24 and π = 6

π β 3π

= 24 β 3(6)

= 24 β 18 = 6

π β π

= 24 β 6 = 18

π + π

= 24 + 6

= 24 + 6 = 30

π + 3π

= 24 + 3(6)

= 24 + 18 = 42

Required numbers are, 6, 18, 30, 42.

If π = 24 and π = β6

π β 3π

= 24 β 3(β6)

= 24 + 18 = 42

π β π

= 24 β (β6)

= 24 + 6 = 30

π + π

= 24 + (β6)

= 24 β 6 = 18

π + 3π

= 24 + 3(β6)

= 24 β 18 = 6

Required numbers are, 42, 30, 18, 6.

**Question **5. Find five numbers in A.P. whose sum is 12 (1/2) and the ratio of the first to the last terms is 2: 3.

Solution:

Let us assumed that five numbers in A.P.,

(π β 2π), (π β π), π, (π + π) πππ (π + 2π)

(π β 2π) + (π β π) + π + (π + π) + (π + 2π) = 12 (1/2)

π β 2π + π β π + π + π + π + π + 2π = 25/2

5π = 25/2

π = 25/2Γ5

π = 5/2

According to question,

πβ2π/π+2π = 2/3

By cross multiplication,

3(π β 2π) = 2(π + 2π)

3π β 6π = 2π + 4π

3π β 2π = 4π + 6π

π = 10π** __**(π)

**Question **6. Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.

Solution:

Let us assumed that three parts in A.P. (π β π), π and (π + π)

(π β π) + π + (π + π) = 207

π β π + π + π + π = 207

3π = 207

π = 207/3

π = 69

According to question,

(π β π) Γ π = 4623

Put the value of a in above equation,

(69 β π) Γ 69 = 4623

(69 β π) = 4623/69

(69 β π) = 67

βπ = 67 β 69

βπ = β2

π = 2

Required numbers are,

π β π = 69 β 2 = 67

π = 69

π + π = 69 + 2 = 71

Hence, the required three parts in A.P. 67, 69 and 71.

**Question **7. The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers.

Solution:

Let us assumed that three parts in A.P. (π β π), π and (π + π)

(π β π) + π + (π + π) = 15

π β π + π + π + π = 15

3π = 15

π = 15/3

π = 5

According to question,

(π β π)^{2} + (π + π)^{2} = 58

(π^{2} + π^{2} β 2ππ) + (π^{2} + π^{2} + 2ππ) = 58

π^{2} + π2 β 2ππ + π^{2} + π^{2} + 2ππ = 58

2π^{2} + 2π^{2} = 58

2(π^{2} + π^{2}) = 58

π^{2} + π^{2} = 58/2

π^{2} + π^{2} = 29

Put the value of a in above equation,

(5)^{2} + π^{2} = 29

25 + π^{2} = 29

π^{2} = 29 β 25

π^{2} = 4

π = β4

π = Β±2

Required numbers are,

If π = 5 and π = 2

π β π = 5 β 2 = 3

π = 5

π + π = 5 + 2 = 7

If π = 5 and π = β2

π β π = 5 β (β2) = 7

π = 5

π + π = 5 + (β2) = 3

Hence, the required three parts in A.P.3,5,7 or 7,5,3.

**Question **8. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Solution:

Let us assumed that four number in A.P. are,

(π β 3π), (π β π), (π + π) and (π + 3π)

(π β 3π) + (π β π) + (π + π) + (π + 3π) = 20

π β 3π + π β π + π + π + π + 3π = 20

4π = 20

π = 20/4

π = 5

According to question,

(π β 3π)^{2 }+ (π β π)^{2} + (π + π)^{2} + (π + 3π)^{2} = 120

π^{2} + (3π)^{2} β 2(π)(3π) + π^{2 }+ π^{2} β 2(π)(π) + π^{2} + π^{2} + 2(π)(π) + π^{2} + (3π)^{2} + 2(π)(3π) = 120

π^{2} + 9π^{2} β 6ππ + π^{2} + π^{2} β 2ππ + π^{2} + π^{2} + 2ππ + π^{2} + 9π^{2} + 6ππ = 120

4π^{2} + 20π^{2} = 120

Take common,

4(π^{2} + 5π^{2}) = 120

π^{2} + 5π^{2} = 120/4

π^{2} + 5π^{2} = 30

Put the value of a in above equation,

(5)^{2} + 5π^{2} = 30

25 + 5π^{2} = 30

5π^{2} = 30 β 25

5π^{2} = 5

π^{2} = 5/5

π^{2} = 1

π = β1

π = Β±1

Required numbers are,

If π = 5 and π = 1

π β 3π = 5 β 3(1) = 5 β 3 = 2

π β π = 5 β 1 = 4

π + π = 5 + 1 = 6

π + 3π = 5 + 3(1) = 5 + 3 = 8

If π = 5 and π = β2

π β 3π = 5 β 3(β1) = 5 + 3 = 8

π β π = 5 β (β1) = 5 + 1 = 6

π + π = 5 + (β1) = 5 β 1 = 4

π + 3π = 5 + 3(β1) = 5 β 3 = 2

Hence, the required three parts in A.P.2,4,6,8 or 8,6,4,2.

**Question **9. Insert one arithmetic mean between 3 and 13.

Solution:

We know that,

Arithmetic mean = π+π/2

So the Arithmetic mean between 3 and 13,

= 3+13/2

= 16/2

= 8

Hence, the Arithmetic mean between 3 and 13 is 8

**Question **10. The angles of a polygon are in A.P. with common difference 5Β°. If the smallest angle is 120Β°, find the number of sides of the polygon.

Solution:

Let us assumed that,

Number of sides of a polygon be n.

It is given that,

Smallest angle π = 120Β°

Common difference π = 5Β°

Sum of interior angles = (2n – 4) Γ 90Λ

π/2 [2 Γ 120Β° + (π β 1) Γ 5Β°] = (2π β 4) Γ 90Β°

π/2 [240 + 5π β 5Β°] = 180π β 360Β°

π[235Β° + 5π] = 360π β 720Β°

235π + 5π^{2} = 360π β 7200

5π^{2}β 125π + 720 = 0

π^{2}β 25π + 144 = 0

π^{2} β 16πβ 9π + 144 = 0

π(π β 16) β 9(π β 16) = 0

(π β 16) (π β 9) = 0

π = 16 ππ π = 9

Hence, the number of side in a polygon is 16 or 9.

**Question **12. Insert four A.M.s between 14 and -1.

Solution:

Let us assumed that, four A. M.s between 14 and -1,

14, π΄1, π΄2, π΄3, π΄4, β1

1^{π π‘} term = 14

π = 14

6^{π‘β} term = β1

π + (6 β 1)π = β1

Put the value of a in above equation,

14 + (6 β 1)π = β1

14 + 5π = β1

5π = β1 β 14

5π = β15

π = β15/5

π = β3

Terms between 14 to -1 are,

π΄_{1 }= 14 + (β3) = 14 β 3 = 11

π΄_{2 }= 14 + 2(β3) = 14 β 6 = 8

π΄_{3} = 14 + 3(β3) = 14 β 9 = 5

π΄_{4} = 14 + 4(β3) = 14 β 12 = 2

Hence, four A.M.s between 14 and -1 are 11, 8, 5 and 2

**Question **13. Insert five A.M.s between -12 and 8.

Solution:

Let us assumed that, five A.M.s between -12 and 8,

β12, π΄1, π΄2, π΄3, π΄4, π΄5, 8

1^{π π‘} term = β12

π = β12

7^{π‘β} term = 8

π + (7 β 1)π = 8

Put the value of a in above equation,

β12 + (7 β 1)π = 8

β12 + 6π = 8

6π = 8 + 12

6π = 20

π = 20/6

π = 10/3

**Question **14. Insert six A.M.s between 15 and -15.

Solution:

Let us assumed that, six A.M.s between 15 and -15,

1_{5}, π΄^{1}, π΄_{2}, π΄_{3}, π΄_{4}, π΄_{5}, π΄_{6}, β15

1^{π π‘} term = 15

π = 15

8^{π‘β} term = β15

π + (8 β 1)π = β15

Put the value of a in above equation,

15 + (8 β 1)π = β15

15 + 7π = β15

7π = β15 β 15

7π = β30

π = β30/7

### Exercise 10 E

**Question **1. Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h^{-1} .The second car goes at a speed of 8 km h^{-1} in the first hour and thereafter increasing the speed by 0.5 km h^{-1} each succeeding hour. After how many hours will the two cars meet?

Solution:

Let us assumed that,

The two cars meet after π₯ hours.

That means the two cars travel the same distance in n hours.

Total Distance covered by 1^{st} car in π₯ hours = 10 Γ π₯ km.

Total Distance covered by 2^{nd} car in π₯ hours =π₯/2 [2 Γ 8 + (π₯ β 1) Γ 0.5]km

10 Γ π₯ = π₯/2 [2 Γ 8 + (π₯ β 1) Γ 0.5]

10Γπ₯Γ2/π₯ = 16 + (π₯ β 1) Γ 0.5

20 = 16 + (π₯ β 1) Γ 0.5

20 β 16 = 0.5π₯β 0.5

4 = 0.5π₯β 0.5

4 + 0.5 = 0.5π₯

0.5π₯ = 4.5

π₯ = 9

Hence, both cars will meet after 9 hours.

**Question **2. A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes

Solution:

It is given that,

Total amount distributed as prize = π_{π} = π
π . 700

Total number of prize = π = 7

Value of prize preceding = -20

Let us assumed that,

First prize will be π₯ = a

We know that,

160 = π

π = 160

Value of 1^{st} Price = 160

Value of 2^{nd} Price = 160 β 20 = 140

Value of 3^{nd} Price = 140 β 20 = 120

Value of 4^{nd} Price = 120 β 20 = 100

Value of 5^{nd }Price = 100 β 20 = 80

Value of 6^{nd} Price = 80 β 20 = 60

Value of 7^{nd }Price = 60 β 20 = 40

Hence, the value of prices is 160, 140, 120, 100, 80, 60 and 40.

**Question **3. An article can be bought by paying Rs. 28,000 at once or by making 12 monthly instalments. If the first instalment paid is Rs. 3,000 and every other instalment is Rs. 100 less than the previous one, find:

(i) amount of instalment paid in the 9^{th} month

(ii) total amount paid in the instalment scheme.

Solution:

It is given that,

Total number of instalments = 12,

First instalment = Rs. 3,000

Amount of instalment decreased = -100

We have,

π = 12

π = 3000

π = β100

(i) amount of instalment paid in the 9^{th} month

We know that,

π = 3000

π = β100

π = 9

ππ = π + (π β 1)π

ππ = 3000 + (9 β 1)(β100)

ππ = 3000 + (8)(β100)

ππ = 3000 β 800

ππ = 2200

Hence, amount paid in 9^{th} month is Rs. 2200.

(ii) total amount paid in the instalment scheme.

π = 12

π = 3000

π = β100

π_{π} = 12/2 [2(3000) + (12 β 1)(β100)]

π_{π} = 6[6000 + 11(β100)]

π_{π} = 6[6000 β 1100]

π_{π} = 6[4900]

π_{π} = 29,400

Hence, the total amount of instalments is Rs. 29,400.

**Question **4. A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7^{th} year. Assuming that the production increases uniformly by a fixed number every year, find:

(i) the production in the first year.

(ii) the production in the 10th year.

(iii) the total production in 7 years.

Solution:

It is given that,

A manufacturer of TV sets produces

3^{rd} year = 600

7^{th} year = 700

Here,

(i) the production in the first year

3^{rd} term = 600

π + (3 β 1)π = 600

π + 2π = 600__________(π)

7^{th }term = 700

π + (7 β 1)π = 700

π + 6π = 700__________(ππ)

From equation (i) we get,

π + 2π = 600

π = 600 β 2π** __**(πππ)

Put the value of a in equation (ii),

600 β 2π + 6π = 700

β2π + 6π = 700 β 600

4π = 100

π = 100/4

π = 25

Put the value of d in equation (iii)

π = 600 β 2(25)

π = 600 β 50

π = 550

(ii) the production in the 10

^{th}year.

π

_{π}= π + (10 β 1)π

Put the values,

π

_{10}= 550 + (9)(25)

π

_{10}= 550 + 225

π

_{10}= 775

Hence, the production in the 10

^{th}year is 775.

**Question **5. Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying instalments every month. If the instalment for the first month is Rs. 1,000 and it increases by Rs. 100 every month, what amount will she pay as the 30^{th} instalment of loan? What amount of loan she still has to pay after the 30^{th} instalment?

Solution:

It is given that,

Total amount of loan = Rs. 1,18,000

First instalment = Rs. 1,000

Increased in instalment = Rs. 100

We know that,

π_{π} = 1,18,000

π = 1,000

π = π
π . 100

Amount of 30^{th} instalment,

π_{π} = π + (π β 1)π

π_{30} = 1000 + (30 β 1)(100)

π_{30} = 1000 + 29(100)

π_{30} = 1000 + 2900

π_{30} = 3900

Total amount paid in 30 instalment,

π_{π} = 15[2000 + (29)100]

π_{π} = 15[2000 + 2900]

π_{π} = 15[4900]

π_{π} = 73,500

Total amount of loan paid after 30^{th} instalment,

Total amount of Loan β Total amount paid

Rs. 1,18,000 β Rs. 73,500

Rs. 44,500

Hence, Total amount of loan paid after 30th instalment is Rs. 44,500

**Question **6. 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 five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?

Solution:

It is given that,

The number of trees, that each section of each class will plant, will be five times of the class to which the

respective section belongs. There are 1 to 10 classes in the school.

Required AP is 1, 2, 3, 4, β¦β¦β¦β¦. 10

π = 1

π = 1

Total number of section = 3

Number of Plants planted by each class = 5

Total number of plants planted by all class = 3 Γ 5 Γ 55

Total number of plants planted by all class = 825

Hence, Total number of plants planted by all class is 825.

### Exercise 10 F

**Question **1. The 6^{th} term of an A.P. is 16 and the 14^{th} term is 32. Determine the 36^{th} term.

Solution:

It is given that,

π_{6} = 16

π + (6 β 1)π = 16

π + 5π = 16 β¦ β¦ β¦ β¦ (π)

And,

π_{14 }= 32

π + (14 β 1)π = 16

π + 13π = 32 β¦ β¦ β¦ β¦ (ππ)

From equation (i) we get

π + 5π = 16

π = 16 β 5π β¦ β¦ β¦ β¦ (πππ)

Put the value of a in equation (ii)

16 β 5π + 13π = 32

β5π + 13π = 32 β 16

8π = 16

π = 16/8

π = 2

Put the value of d in equation (iii)

π = 16 β 5(2)

π = 16 β 10

π = 6

To find 36th term,

π_{36} = π + (36 β 1)π

π_{36} = 6 + (36 β 1)2

π_{36} = 6 + (35)2

π_{36} = 6 + 70

π_{36} = 76

**Question **2. If the third and the 9^{th} terms of an A.P. term is 4 and -8. Find the general term. Which term of AP is 0.

Solution:

It is given that,

π_{3} = 4

π + (3 β 1)π = 4

π + 2π = 4 β¦ β¦ β¦ β¦ (π)

And,

π_{9} = β8

π + (9 β 1)π = β8

π + 8π = β8 β¦ β¦ β¦ β¦ (ππ)

From equation (i) we get the value of a,

π = 4 β 2π β¦ β¦ β¦ β¦ (πππ)

Put the value of a in equation (ii)

4 β 2π + 8π = β8

6π = β8 β 4

6π = β12

π = β12/6

π = β2

Put the value of d in equation (iii)

π = 4 β 2π

π = 4 β 2(β2)

π = 4 + 4

π = 8

To find General term =

π_{π} = 8 + (π β 1)(β2)

π_{π} = 8 + β2π + 2

π_{π} = 10 β 2π

Let us assumed that,

π^{π‘β} term of this A.P. be 0

8 + (πβ 1)(β2) = 0

8β 2π + 2 = 0

10β 2π = 0

β2π = β10

π = 5

Hence, the 5^{th} term of the AP is 0.

**Question **3. An A.P. consists of 50 terms of which 3^{rd} term is 12 and the last term is 106. Find the 29^{th} term of the A.P.

Solution:

It is given that,

Total number of term is 50,

3^{rd} π‘πππ = 12

π + 2π = 12 β¦ β¦ . . β¦ . (π)

50^{π‘β} π‘πππ = 106

π + 49π = 106 β¦ β¦ β¦ β¦ . (ππ)

From equation (i) we get,

π = 12 β 2π β¦ β¦ β¦ β¦ . (πππ)

Put the value of a in equation (ii)

12 β 2π + 49π = 106

β2π + 49π = 106 β 12

47π = 94

π = 94/47

π = 2

Put the value of d in equation (iii)

π = 12 β 2(2)

π = 12 β 4

π = 8

To find 29^{th} term,

π_{29} = 8 + (29 β 1)(2)

π_{29} = 8 + (28)(2)

π_{29} = 8 + 56

π_{29} = 64

Hence, the value of 29^{th} term is 64.

**Question **4. Find the arithmetic mean of:

(i) -5 and 41

(ii) 3π₯β 2π¦ and 3π₯ + 2π¦

(iii) (π + π)^{2} and (π β π)^{2}

Solution:

(i) -5 and 41

The Arithmetic mean of -5 and 41,

β5+41/2

36/2 = 18

Hence, the arithmetic mean of -5 and 41 is 18

**Question **5. Find the sum of first 10 terms of the A.P. 4 + 6 + 8 + β¦β¦β¦

Solution:

It is given that,

π = 4

π = 6β 4

π = 2

π = 10

To the find Sum of 10^{th} terms,

**Question **6. Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.

Solution:

It is given that,

π = 3

π = 57

π = 20

To the find Sum of 20^{th} terms,

**Question **7. How many terms of the series 18 + 15 + 12 + β¦β¦. when added together will give 45?

Solution:

It is given that,

The AP is 18, 15, 12,β¦.β¦β¦.

π = 18

π = 15 β 18

π = β3

π_{π} = 45

Put the values,

90 = π[36β 3π + 3]

90 = π[39β 3π]

90 = 3π[13β π]

30 = 13πβ π^{2}

π^{2}β 13π + 30 = 0

π^{2}β 10π β 3π + 30 = 0

π(π β 10) β 3(π β 10) = 0

(π β 10)(π 3β ) = 0

π β 10 = 0, π β 3 = 0

π = 10, π = 3

Hence, required number of term to be added is 3 or 10.

**Question **8. The nth term of a sequence is 8 β 5n. Show that the sequence is an A.P.

Solution:

It is given that,

π‘_{π} = 8β 5π

Replace the value of n by (n + 1), we get

π‘_{π}+1 = 8β 5(π + 1)

π‘_{π}+1 = 8β 5πβ 5

π‘_{π}+1 = 3β 5π

Subtract equation (ii) from (i)

π‘(π + 1) β π‘_{π}

= (3β 5π)β (8β 5π)

= 3β 5πβ 8 + 5π

= β5

As, (π‘_{π} + 1 β π‘_{π}) is a constant, Hence, the sequence is an A.P.

**Question **9. The the general term (n^{th} term) and 23r^{d} term of the sequence 3, 1, -1, -3, β¦β¦

Solution:

It is given that,

The AP is

3, 1, -1, -3, β¦..

π = 3

π = 1 β 3

π = β2

We know that,

π_{π} = π + (πβ 1)π

π_{π} = 3 + (πβ 1)(β2)

π_{π} = 3β 2π + 2

π_{π} = 5β 2π

To find 23^{rd} term,

π_{π} = π + (πβ 1)π

π_{23} = 3 + (23β 1)(β2)

π_{23} = 3 + (22)(β2)

π_{23} = 3 β 44

π_{23} = β41

Hence, the 23^{rd} term is -41.

**Question **10. Which term of the sequence 3, 8, 13, β¦β¦β¦β¦.. is 78 ?

Solution:

It is given that,

The AP is

3, 8, 13,..β¦β¦..78

π = 3

π = 8 β 3

π = 5

π_{π} = 78

π =?

π_{π} = π + (πβ 1)π

78 = 3 + (πβ 1)(5)

78 β 3 = (πβ 1)(5)

75 = 5πβ 5

75 + 5 = 5π

5π = 80

π = 16

Hence, 78 is the 16^{th} term of given AP.

**Question **11. Is -150 a term of 11, 8, 5, 2,β¦β¦β¦β¦?

Solution:

It is given that,

The AP is

11, 8, 5, 2, β¦β¦β¦β¦..

π = 11

π = 8 β 11

π = β3

π_{π} = β150

π =?

π_{π} = π + (πβ 1)π

β150 = 11 + (π β 1)(β5)

β150 β 11 = (π β 1)(β5)

β161 = β5π + 5

5_{π} = 166

π = 166/5

π canβt be a fraction.

Hence, -150 is not a term of AP.

**Question **12. How many two digit numbers are divisible by 3 ?

Solution:

We know that,

Smallest 2 digit number be 12 which is divisible by 3.

Largest 2 digit number be 99 which is divisible by 3.

So the required AP is

12, 15, 18, 21,β¦β¦β¦β¦β¦β¦.. 99

π = 12

π = 15 β 12

π = 3

π_{π} = 99

π_{π} = π + (πβ 1)π

99 = 12 + (πβ 1)(3)

99β 12 = (π β 1)(3)

99 β 12 + 3 = 3π β 3

90 = 3π

π = 30

Hence, the two digit numbers divisible by 3 are 30.

**Question **13. How many multiples of 4 lie between 10 and 250?

Solution:

We know that,

Smallest number between 10 and 250 which is divisible by 4 is 12.

Largest number between 10 and 250 which is divisible by 4 is 248.

So, the required AP is,

12, 16, 20, 24,β¦β¦β¦β¦β¦β¦,248

π = 12

π = 16 β 12

π = 4

π_{π} = 248

π_{π} = π + (πβ 1)π

248 = 12 + (πβ 1)(4)

248 = 12 + 4πβ 4

248 = 8 + 4π

248 β 8 = 4π

240 = 4π

240/4 = π

π = 60

Hence, the multiples of 4 lie between 10 and 250 are 60.

**Question **14. The sum of the 4^{th} term and the 8th term of an A.P. is 24 and the sum of 6^{th} term and the 10^{th} term is 44. Find the first three terms of the A.P.

Solution:

It is given that,

π_{4} + π_{8} = 24

π + (4 β 1)π + π + (8 β 1)π = 24

π + 3π + π + 7π = 24

2π + 10π = 24

2(π + 5π) = 24

π + 5π = 24/2

π + 5π = 12 β¦ β¦ β¦ . . (π)

And,

π_{6} + π_{10} = 44

π + (6 β 1)π + π + (10 β 1)π = 44

(π + 5π) + (π + 9π) = 44

2π + 14π = 44

2(π + 7π) = 44

π + 7π = 44/2

π + 7π = 22 β¦ β¦ β¦ β¦ . (ππ)

From equation (i) we get,

π = 12 β 5π β¦ β¦ β¦ β¦ . (πππ)

Put the value π in equation (ii)

12 β 5π + 7π = 22

2π = 22 β 12

2π = 10

π = 10/2

π = 5

Put the value of π in equation (iii)

π + 25 = 12

π = β13

1st term = β13

2^{nd }term = π + π

= β13 + 5

= β8

3^{rd} term = π + 2π

= β13 + 2 Γ 5

= β13 + 10

= β3

Hence, the first three terms of an A.P. are β13, β8 and β5.

**Question **15. The sum of first 14 terms of an A.P. is 1050 and its 14^{th }term is 140. Find the 20^{th} term**Solution:**It is given that,

π

_{14}= 1050

14/2 [2π + (14 β 1)π] = 1050

7[2π + 13π] = 1050

2π + 13π = 1050/7

2π + 13π = 150 β¦ β¦ β¦ β¦ β¦ β¦ β¦ (π)

Also,

π

_{14}= 140

π + (14 β 1)π = 140

π + 13π = 140 β¦ β¦ β¦ β¦ β¦ β¦ β¦ (ππ)

From equation (ii) we get,

π + 13π = 140

π = 140 β 13π β¦ β¦ β¦ β¦ β¦ β¦ β¦ (πππ)

Put the value of a in equation (i)

2(140 β 13π) + 13π = 150

280 β 26π + 13π = 150

β13π = 150 β 280

β13π = β130

π = 10

Put the value of d in equation (iii)

π = 140 β 13(10)

π = 140 β 130

π = 10

To find 20

^{th }term,

= 10 + 19π

= 10 + 19(10)

= 200

Hence, the value of 20

^{th}term us 200 of the given AP.

**Question **16. The 25^{th} term of an A.P. exceeds its 9^{th} term by 16. Find its common difference.

Solution:

n^{th} term of an A.P. is

It is given

π_{π} = π + (πβ 1)π

π_{π} = π + (25β 1)π

π_{25} = π + 24π

π_{9} = π + (9 β 1)π

π_{9} = π + 8π

As per question,

π_{25} = π_{9} + 16

π + 24π = π + 8π + 16

24π β 8π = 16

16π = 16

π = 1

Hence, the common difference is 1.

**Question **17. For an A.P., show that: (π + π)^{π‘β} π‘πππ + (πβ π)π‘βπ‘πππ = 2 Γ π^{π‘β} π‘πππ

Solution:

(π + π)^{π‘β} π‘πππ

π + (π + πβ 1)π β¦ β¦ β¦ β¦ β¦ (π)

(π β π)^{π‘β }π‘πππ

π + (π β πβ 1)π β¦ β¦ β¦ β¦ β¦ . (ππ)

From (i)+(ii), we get

(π + π)^{π‘β} π‘πππ + (πβ π)^{π‘β} π‘πππ

π + (π + πβ 1)π + π + (π β πβ 1)π

π + ππ + ππ β π + π + ππβ ππβ π

2π + 2ππβ 2π

2π + (πβ 1)2π

2[π + (πβ 1)π]

2 Γ π^{π‘β} π‘πππ

Hence Proved

**Question **18. If the nth term of the A.P. 58, 60, 62,β¦β¦.. is equal to the nth term of the A.P. -2, 5, 12,β¦β¦., find the value of n.

Solution:

It is given that,

58, 60, 62, β¦ β¦ β¦.

π = 58

π = 2

π_{π} = π + (πβ 1)π

π_{π} = 58 + (πβ 1)2 β¦ β¦ β¦ β¦ . (π)

Now,

β2, 5, 12, β¦ β¦ β¦ ..

π = β2

π = 7

π_{π} = π + (πβ 1)π

π_{π} = β2 + (πβ 1)7 β¦ β¦ β¦ β¦ . . (ππ)

From (i) and (ii) we get,

58 + (πβ 1)2 = β2 + (πβ 1)7

58 + 2πβ 2 = β2 + 7πβ 7

56 + 2π = β9 + 7π

56 + 9 = 7π β 2π

65 = 5π

π = 15

**Question **19. Which term of the A.P. 105, 101, 97β¦β¦β¦β¦β¦β¦β¦.. is the first negative term?

Solution:

Let us assumed that,

ππ be the first negative term

π = 105

π = 101 β 105

π = β4

π_{π} < 0

π + (πβ 1)π < 0

105 + (π β 1)(β4) < 0

105 + β4π + 4 < 0

109 β 4π < 0

**Question **20. How many three digit numbers are divisible by 7?

Solution:

We now that,

First three digit number divisible by 7 = 105

Last digit which is divisible by 7 is 994

This is an A.P. in which

π = 105,

π = 7

π_{π} = 994

π_{π} = π + (πβ 1)π

994 = 105 + (π β 1)7

994 β 105 = (π β 1)7

889 = 7πβ 7

889 + 7 = 7π

896 = 7π

896/7

= π

π = 128

Hence, three digit numbers which are divisible by 7 are 128.

**Question **21. Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.

Solution:

It is given that,

The three parts of 216 in A.P

Let us assumed that, the required A.P. is (πβ π), π, (π + π)

πβ π + π + π + π = 216

3π = 216

π = 72

According to question,

Product of the two smaller parts = 5040.

π(πβ π) = 5040

72(72β π) = 5040

72β π = 70

π = 2

πβ π = 72 β 2 = 70

π = 72

π + π = 72 + 2 = 74

Hence, the three parts of 216 are 70, 72 and 74.

**Question **22. Can 2π^{2} β 7 be the nth term of an A.P? Explain.

Solution:

It is given that, the required AP is 2π^{2} β 7,

π = 1

2(1)^{2} β 7

2 Γ 1β 7

2 β 7

β5

π = 2

2(2)^{2} β 7

2 Γ 4 β 7

8 β 7

1

π = 3

2(3)^{2} β 7

2 Γ 9 β 7

18 β 7

11

So, the required AP is β5, 1, 11, β¦ β¦ β¦ .

π = β5

π = 1 β (β5)

π = 1 + 5

π = 6

π = 11 β 1

π = 10

Here, the common difference is not same

Hence, the n^{th} term of an A.P canβt be 2n^{2} β 7.

**Question **23. Find the sum of the A.P., 14, 21, 28,β¦β¦β¦β¦,168.

Solution:

It is given that,

14, 21, 28,β¦β¦β¦β¦,168

π = 14

π = 7

π_{π} = 168

π_{π }= π + (πβ 1)π

168 = 14 + (πβ 1)7

168 β 14 = (πβ 1)7

154 = 7πβ 7

154 = 7πβ 7

161 = 7π

π = 23

**Question **24. The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^{st} term of this A.P.

Solution:

It is given that,

π = 20

π7 = 2100

**Question **25. Find the sum of last 8 terms of the A.P. -12, -10, -8,β¦.β¦β¦,58.

Solution:

It is given that,

β12, β10, β8, β¦ . β¦ β¦ ,58

Reverse the term,

58, β¦ . , β8, β10, β12

π = 58

π = β2

π_{π} = π/2 [2π + (π β 1)π]

π_{8} = 8/2 [2(58) + (8 β 1)(β2)]

π_{8} = 4[116 + (7)(β2)]

π_{8} = 4[116 β 14]

π_{8} = 4[102]

π_{8} = 408

Hence, the sum of last 8 terms of the A.P. -12, -10, -8, β¦β¦, 58 is 408.