# Selina ICSE Class 10 Maths Solutions Chapter 10 Arithmetic Progression

Question 1. Which of the following sequences are in arithmetic progression?
(π) 2, 6, 10, 14,
(ππ) 15, 12, 9, 6,
(πππ) 5, 9, 12, 18,
(ππ£) 1/2, 1/3, 1/4, 1/5
Solution:

(π) 2, 6, 10, 14,
π = π2 β π1
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.
(ππ) 15, 12, 9, 6,
π = π2 β π1
π1 = 12 β 15
π1 = β3
π2 = 9 β 12
π2 = β3
π2 = 6 β 9
π3 = β3
Here,
π1 = π2 = π3
Hence, the given sequence is in arithmetic progression.
(πππ) 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 ππ‘β term of sequence is (2πβ 3), find its fifteenth term.
Solution:

It is given that,
ππ‘β term of sequence is (2πβ 3)
So, 15π‘β term is
2(15)β 3
30 β 3 = 27
Hence, the fifteenth term is 27.

Question 3. If the ππ‘β term of an A.P. is (2π + 3), find the A.P.
Solution:

It is given that,
pπ‘β term of sequence is (2π + 3)
So, 1π π‘ term is
2(1) + 3
2 + 3 =5
So, 1π π‘ term is
2(1) + 3
2 + 3 =5
So, 2ππ term is
2(2) + 3
4 + 3 =7
So, 3ππ term is
2(3) + 3
6 + 3 = 9
Hence, the AP is 5, 7, 9,β¦β¦β¦β¦ .

Question 4. Find the 24π‘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 = β2
π = 12
π = 24
ππ‘β = π + (π β 1)π
24π‘β = 12 + (24 β 1)(β2)
24π‘β = 12 + (23)(β2)
24π‘β = 12 + (23)(β2)
24π‘β = 12 β 46
24π‘β = β34
Hence, the 24π‘β term is β34.

Question 5. Find the 30π‘β term of the sequence:

Question 6. Find the 100π‘β term of the sequence β3, 2β3, 3β3, β¦ β¦ β¦ β¦ β¦ . .. .
Solution:

It is given that,
β3, 2β3, 3β3, β¦ β¦ β¦ β¦ β¦
π = 2β3 β β3
π = β3
π = β3
π = 100π‘β
ππ‘β = π + (π β 1)π
100π‘β = β3 + (100 β 1)β3
100π‘β = β3 + 100β3 β β3
100π‘β = 100β3
Hence, the 100π‘β term of the given AP is 100β3.

Question 7. Find the 50π‘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 6π‘β terms of an A.P are respectively 6 and 5. Find the 11π‘β term of the A.P.
Solution:

We know that,
ππ‘β = π + (π β 1)π
5π‘β = 6
π + (5 β 1)π = 6
π + 4π = 6_____________(π)
6π‘β = 5
π + (6 β 1)π = 5
π + 5π = 5_____________(ππ)
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 = β1
Put the value of d in equation (i)
π + 4(β1) = 6
π β 4 = 6
π = 6 + 4
π = 10
Now,
π = 11
ππ‘β = 10 + (11 β 1)(β1)
ππ‘β = 10 + 10(β1)
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______________(ππ)
From equation (i) we get,
π + 3π = 10
π = 10 β 3π__(πππ)
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 = β3
π = β1
Put the value of d in equation (iii)
π = 10 β 3(β1)
π = 10 + 3
π = 13
Hence, the value of π is 13 and value of π is β1.

Question 14. Find the 10π‘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,
10th 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 3ππ term is 5 and 7π‘β term is 9.
Solution:

It is given that,
3ππ term = 5
7π‘β term = 9
From 3ππ term we get,
3ππ 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______________(ππ)
From equation (i) we get,
π + 2π = 5
π = 5 β 2π__(πππ)
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 31π π‘ term of an A.P whose 10π‘β term is 38 and 16π‘β term is 74.
Solution:

It is given that,
10ππ term = 38
16π‘β term = 74
From 10th term we get,
10π‘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______________(ππ)
From equation (i) we get,
π + 9π = 38
π = 38 β 9π__(πππ)
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 = β16
To find 31st term,
ππ‘β = π + (π β 1)π
31π‘β = β16 + (31 β 1)6
31π‘β = β16 + (30)6
31π‘β = β16 + 180
31π‘β = 164
Hence, the 31st 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 = β3
π = 21
ππ‘β = β81
ππ‘β = π + (π β 1)π
β81 = 21 + (π β 1)(β3)
β81 β 21 = β3π + 3
β102 = β3π + 3
β102 β 3 = β3π
β105 = β3π
3π = 105
π = 105/3
π = 35
Hence, the 35th term of the AP is -81.
Now,
ππ‘β = π + (π β 1)π
0 = 21 + (π β 1)(β3)
0 β 21 = β3π + 3
β21 = β3π + 3
β21 β 3 = β3π
β24 = β3π
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 31st 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 4th and the 8th 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 + π8 = 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 + π10 = 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¦ . (ππ)
From equation (i) we get,
π = 12 β 5π
Put the value of a in equation (ii)
12 β 5π + 7π = 17
β5π + 7π = 17 β 12
2π = 5
π = 5/2
Put the value of βdβ in (iii)

Question 20. If the third term of an A.P. is 5 and the seventh terms are 9, find the 17th 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¦ (ππ)
From equation (i), we get
π + 2π = 5
π = 5 β 2π
Put the value a in equation (ii)
5 β 2π + 6π = 9
β2π + 6π = 9 β 5
4π = 4
π = 1
Put the value of d in equation (i)
π + 2(1) = 5
π = 3
Now,
17π‘β = π + (π β 1)π
17π‘β = 3 + (17 β 1)(1)
17π‘β = 3 + 16 Γ 1
17π‘β = 19
Hence, the value of 31st term is 67.

### Exercise 10B

Question 1. In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th 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
β2π = 78π
π = β78π/2
π = β39π
To find π40
π40 = β39π + (40 β 1)π
π40 = β39π + 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 31st 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 βπ31 = 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,
3ππ π‘πππ = 4
π + (3 β 1)π = 4
π + 2π = 4 β¦ β¦ β¦ β¦ β¦ (π)
9π‘β π‘πππ = β8
π + (9 β 1)π = β8
π + 8π = β8 β¦ β¦ β¦ β¦ β¦ (ππ)
From equation (i) we get,
π = 4 β 2π β¦ β¦ β¦ β¦ β¦ (πππ)
Put the value of a in equation (ii)
4 β 2π + 8π = β8
6π = β8 β 4
6π = β12
π = β12/6
π = β2
Put the value of d in equation (iii)
π = 4 β 2(β2)
π = 4 + 4
π = 8
Let us assume that,
ππ‘β = π + (π β 1)π
0 = 8 + (π β 1)(β2)
β8 = β2π + 2
β8 β 2 = β2π
β10 = β2π
β10/β2 = π
π = 5
Hence, 5th 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‘πππ = 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 = β4 β 61
β5π = β65
π = β65/β5
π = 13
Hence, the value of n is 13.

Question 8. Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:

It is given that,
3ππ π‘πππ = 16
π + (3 β 1)π = 16
π + 2π = 16_______________(π)
Also,
7π‘β π‘πππ β 5π‘β π‘πππ = 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 πβ 2,4πβ 1 and 5π + 2 are in A.P. find the value of n and its next two terms.
Solution:

It is given that,
πβ 2, 4πβ 1 and 5π + 2 are in A.P.
(4πβ 1) β (πβ 2) = (5π + 2) β (4πβ 1)
4πβ 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
= β1
Second term = 4πβ 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¦.
4π‘β π‘πππ = π + (π β 1)π
= β1 + (4 β 1)4
= β1 + (3)4
= β1 + 12
= 11
5π‘β π‘πππ = π + (π β 1)π
= β1 + (5 β 1)4
= β1 + (4)4
= β1 + 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
βπ β π = β2 + 2
β2π = 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 7π‘β term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution:

It is given that,
Total number of terms is 57.
7π‘β π‘πππ = 13
π + (7 β 1)π = 13
π + 6π = 13_______________(π)
57π‘β π‘πππ = 108
π + (57 β 1)π = 108
π + 56π = 108_______________(ππ)
From equation (i) we get,
π = 13 β 6π_(πππ)
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. 4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd time by 1. Find the first term and the common difference.
Solution:

It is given that,
4π‘β π‘πππ = 3 Γ 1π π‘ π‘πππ
π + (4 β 1)π = 3π
π + 3π = 3π
π β 3π + 3π = 0
2π β 3π = 0________________(π)
Again,
7π‘β π‘πππ β 2(3ππ π‘πππ) = 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
βπ + 2π = 1________________(ππ)
From equation (i) get the value of a,
2π = 3π

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

It is given that,
2π‘β π‘πππ + 7π π‘ π‘πππ = 30
π + (2 β 1)π + π + (7 β 1)π = 30
π + π + π + 6π = 30
2π + 7π = 30______________(π)
Again,
2 Γ 8π‘β π‘πππ β 15π π‘ π‘πππ = 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 mth term is n and nth term is m, show that its rth term is (m + n β r)
Solution:

It is given that,
ππ‘β π‘πππ = π
π + (π β 1)π = π(π)
ππ‘β π‘πππ = π
π + (π β 1)π = π(ππ)
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 ππ + π = π β π
ππ β ππ = π β π
π(π β π) = π β π
βπ(π β π) = π β π
βπ = πβπ/πβπ
π = β1
Put the value of d in equation (i)
π + (π β 1)(β1) = π
π β π + 1 = π
π = π + π β 1
According to question,
ππ‘β π‘πππ
π + (π β 1)π
(π + π β 1) + (π β 1)(β1)
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 13th term?
Solution:

It is given that,
3, 10, 17, β¦ β¦ β¦ β¦
π = 3
π = 10 β 3
π = 7
To find 13th term,
13π‘β π‘πππ = 3 + (13 β 1)7
13π‘β π‘πππ = 3 + (12)7
13π‘β π‘πππ = 3 + 84
13π‘β π‘πππ = 87
According to question,
ππ‘β π‘πππ β 13π‘β π‘πππ = 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, β2, β¦ β¦ β¦ ..
Solution:

It is given that,
8, 3, β2, β¦ β¦ β¦ . .
π = 8
π = 3 β 8
π = β5
π = 22
ππ = π/2 [2π + (π β 1)π]
π22 = 22/2[2(8) + (22 β 1)(β5)]
π22 = 11[16 + 21(β5)]
π22 = 11[16 β 105]
π22 = 11 Γ (β89)
π22 = β979
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 = β3
ππ = 78
ππ = π/2 [2π + (π β 1)π]
78 = π/2 [2(24) + (π β 1)(β3)]
78 Γ 2 = π[48 + (π β 1)(β3)]
156 = π[48 + (π β 1)(β3)]
156 = π[48 + (β3π + 3)]
156 = π[48 β 3π + 3]
156 = π[51 β 3π]
156 = 51π β 3π2
51π β 3π2 β 156 = 0
β3(β17π + 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 8πβ 5.
Solution:

It is given that,
nth term of an AP is 8πβ 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 2nd and 3rd terms are 14 and 18 respectively.
Solution:

πΌπ‘ ππ  πππ£ππ π‘βππ‘,
2ππ π‘πππ = 14
π + (2 β 1)π = 14
π + π = 14_________________(π)
3ππ π‘πππ = 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 = β17
Sum of n terms = 132

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

It is given that,
8π‘β π‘πππ = 37
π + (8 β 1)π = 37
π + 7π = 37_____________(π)
15π‘β π‘πππ β 12π‘β π‘πππ = 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 8th term.
Solution:

It is given that,
Sum of n natural numbers is 5π2 + 4π
ππ = 5π2 + 4π
ππβ1 = 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‘πππ = ππ β ππβ1
= 5π2 + 4π β (5π2 β 6π + 1)
= 5π2 + 4π β 5π2 + 6π β 1
= 10π β 1
To find 8th 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,
4π‘β π‘πππ = 11
π + (4 β 1)π = 11
π + 3π = 11____________(π)
Also,
8π‘β π‘πππ β 2(4π‘β π‘πππ) = 5
π + (8 β 1)π β 2(11) = 5
π + 7π β 22 = 5
π + 7π = 5 + 22
π + 7π = 27____________(ππ)
From equation (i) we get,
π + 3π = 11
π = 11 β 3π_(πππ)
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 = β1
First term = β1
Second term = β1 + 4 = 3
Third term = β1 + 2(4) = β1 + 8 = 7
So, the required AP is β1,3,7, β¦ β¦ β¦ β¦ β¦ β¦.
Sum of 50 terms,
ππ = π/2 [2π + (π β 1)π]
π50 = 50/2 [2(β1) + (50 β 1)4]
π50 = 25[β2 + (49)4]
π50 = 25[β2 + 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 = β9
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 = β3
8 β (β3),8 and 8 + (β3)
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 2ππ + π2 + π2 + π2 + 2ππ + 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 = β9
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 = β3
7 β (β3),9 and 7 + (β3)
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° πππ 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) πππ (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 = β36
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 = β6
π β 3π
= 24 β 3(β6)
= 24 + 18 = 42
π β π
= 24 β (β6)
= 24 + 6 = 30
π + π
= 24 + (β6)
= 24 β 6 = 18
π + 3π
= 24 + 3(β6)
= 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) πππ (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 = 25/2Γ5
π = 5/2
According to question,
πβ2π/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
βπ = 67 β 69
βπ = β2
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 2ππ) + (π2 + π2 + 2ππ) = 58
π2 + π2 β 2ππ + π2 + π2 + 2ππ = 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 = β4
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 = β2
π β π = 5 β (β2) = 7
π = 5
π + π = 5 + (β2) = 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 6ππ + π2 + π2 β 2ππ + π2 + π2 + 2ππ + π2 + 9π2 + 6ππ = 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 = β1
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 = β2
π β 3π = 5 β 3(β1) = 5 + 3 = 8
π β π = 5 β (β1) = 5 + 1 = 6
π + π = 5 + (β1) = 5 β 1 = 4
π + 3π = 5 + 3(β1) = 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2β 125π + 720 = 0
π2β 25π + 144 = 0
π2 β 16πβ 9π + 144 = 0
π(π β 16) β 9(π β 16) = 0
(π β 16) (π β 9) = 0
π = 16 ππ 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, β1
1π π‘ term = 14
π = 14
6π‘β term = β1

π + (6 β 1)π = β1
Put the value of a in above equation,
14 + (6 β 1)π = β1
14 + 5π = β1
5π = β1 β 14
5π = β15
π = β15/5
π = β3
Terms between 14 to -1 are,
π΄1 = 14 + (β3) = 14 β 3 = 11
π΄2 = 14 + 2(β3) = 14 β 6 = 8
π΄3 = 14 + 3(β3) = 14 β 9 = 5
π΄4 = 14 + 4(β3) = 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,
β12, π΄1, π΄2, π΄3, π΄4, π΄5, 8
1π π‘ term = β12
π = β12
7π‘β term = 8
π + (7 β 1)π = 8
Put the value of a in above equation,
β12 + (7 β 1)π = 8
β12 + 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,
15, π΄1, π΄2, π΄3, π΄4, π΄5, π΄6, β15
1π π‘ term = 15
π = 15
8π‘β term = β15
π + (8 β 1)π = β15
Put the value of a in above equation,
15 + (8 β 1)π = β15
15 + 7π = β15
7π = β15 β 15
7π = β30
π = β30/7

### 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 1st car in π₯ hours = 10 Γ π₯ km.
Total Distance covered by 2nd 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]
10Γπ₯Γ2/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 1st Price = 160
Value of 2nd Price = 160 β 20 = 140
Value of 3nd Price = 140 β 20 = 120
Value of 4nd Price = 120 β 20 = 100
Value of 5nd Price = 100 β 20 = 80
Value of 6nd Price = 80 β 20 = 60
Value of 7nd 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 9th 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 = β100
(i) amount of instalment paid in the 9th month
We know that,
π = 3000
π = β100
π = 9
ππ = π + (π β 1)π
ππ = 3000 + (9 β 1)(β100)
ππ = 3000 + (8)(β100)
ππ = 3000 β 800
ππ = 2200
Hence, amount paid in 9th month is Rs. 2200.
(ii) total amount paid in the instalment scheme.
π = 12
π = 3000
π = β100
ππ = 12/2 [2(3000) + (12 β 1)(β100)]
ππ = 6[6000 + 11(β100)]
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 7th 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
3rd year = 600
7th year = 700
Here,
(i) the production in the first year
3rd term = 600
π + (3 β 1)π = 600
π + 2π = 600__________(π)
7th term = 700
π + (7 β 1)π = 700
π + 6π = 700__________(ππ)
From equation (i) we get,
π + 2π = 600
π = 600 β 2π__(πππ)
Put the value of a in equation (ii),
600 β 2π + 6π = 700
β2π + 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 10th 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 10th 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 30th instalment of loan? What amount of loan she still has to pay after the 30th 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 30th 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 30th 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 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th 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¦ (ππ)
From equation (i) we get
π + 5π = 16
π = 16 β 5π β¦ β¦ β¦ β¦ (πππ)
Put the value of a in equation (ii)
16 β 5π + 13π = 32
β5π + 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 9th 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 = β8
π + (9 β 1)π = β8
π + 8π = β8 β¦ β¦ β¦ β¦ (ππ)
From equation (i) we get the value of a,
π = 4 β 2π β¦ β¦ β¦ β¦ (πππ)
Put the value of a in equation (ii)
4 β 2π + 8π = β8
6π = β8 β 4
6π = β12
π = β12/6
π = β2
Put the value of d in equation (iii)
π = 4 β 2π
π = 4 β 2(β2)
π = 4 + 4
π = 8
To find General term =
ππ = 8 + (π β 1)(β2)
ππ = 8 + β2π + 2
ππ = 10 β 2π
Let us assumed that,
ππ‘β term of this A.P. be 0
8 + (πβ 1)(β2) = 0
8β 2π + 2 = 0
10β 2π = 0
β2π = β10
π = 5
Hence, the 5th term of the AP is 0.

Question 3. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:

It is given that,
Total number of term is 50,
3rd π‘πππ = 12
π + 2π = 12 β¦ β¦ . . β¦ . (π)
50π‘β π‘πππ = 106
π + 49π = 106 β¦ β¦ β¦ β¦ . (ππ)
From equation (i) we get,
π = 12 β 2π β¦ β¦ β¦ β¦ . (πππ)
Put the value of a in equation (ii)
12 β 2π + 49π = 106
β2π + 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 29th term,
π29 = 8 + (29 β 1)(2)
π29 = 8 + (28)(2)
π29 = 8 + 56
π29 = 64
Hence, the value of 29th 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,
β5+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 10th 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 20th 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 = β3
ππ = 45
Put the values,

90 = π[36β 3π + 3]
90 = π[39β 3π]
90 = 3π[13β π]
30 = 13πβ π2
π2β 13π + 30 = 0
π2β 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 5πβ 5
π‘π+1 = 3β 5π
Subtract equation (ii) from (i)
π‘(π + 1) β π‘π
= (3β 5π)β (8β 5π)
= 3β 5πβ 8 + 5π
= β5
As, (π‘π + 1 β π‘π) is a constant, Hence, the sequence is an A.P.

Question 9. The the general term (nth term) and 23rd 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 = β2
We know that,
ππ = π + (πβ 1)π
ππ = 3 + (πβ 1)(β2)
ππ = 3β 2π + 2
ππ = 5β 2π
To find 23rd term,
ππ = π + (πβ 1)π
π23 = 3 + (23β 1)(β2)
π23 = 3 + (22)(β2)
π23 = 3 β 44
π23 = β41
Hence, the 23rd 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 + (πβ 1)π
78 = 3 + (πβ 1)(5)
78 β 3 = (πβ 1)(5)
75 = 5πβ 5
75 + 5 = 5π
5π = 80
π = 16
Hence, 78 is the 16th 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 = β3
ππ = β150
π =?
ππ = π + (πβ 1)π
β150 = 11 + (π β 1)(β5)
β150 β 11 = (π β 1)(β5)
β161 = β5π + 5
5π = 166
π = 166/5
π canβt 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 + (πβ 1)π
99 = 12 + (πβ 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 + (πβ 1)π
248 = 12 + (πβ 1)(4)
248 = 12 + 4πβ 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 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th 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¦ . (ππ)
From equation (i) we get,
π = 12 β 5π β¦ β¦ β¦ β¦ . (πππ)
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 = β13
1st term = β13
2nd term = π + π
= β13 + 5
= β8
3rd term = π + 2π
= β13 + 2 Γ 5
= β13 + 10
= β3
Hence, the first three terms of an A.P. are β13, β8 and β5.

Question 15. The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th 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¦ (ππ)
From equation (ii) we get,
π + 13π = 140
π = 140 β 13π β¦ β¦ β¦ β¦ β¦ β¦ β¦ (πππ)
Put the value of a in equation (i)
2(140 β 13π) + 13π = 150
280 β 26π + 13π = 150
β13π = 150 β 280
β13π = β130
π = 10
Put the value of d in equation (iii)
π = 140 β 13(10)
π = 140 β 130
π = 10
To find 20th term,
= 10 + 19π
= 10 + 19(10)
= 200
Hence, the value of 20th term us 200 of the given AP.

Question 16. The 25th term of an A.P. exceeds its 9th term by 16. Find its common difference.
Solution:

nth term of an A.P. is
It is given
ππ = π + (πβ 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‘πππ + (πβ π)π‘βπ‘πππ = 2 Γ ππ‘β π‘πππ
Solution:

(π + π)π‘β π‘πππ
π + (π + πβ 1)π β¦ β¦ β¦ β¦ β¦ (π)
(π β π)π‘β π‘πππ
π + (π β πβ 1)π β¦ β¦ β¦ β¦ β¦ . (ππ)
From (i)+(ii), we get
(π + π)π‘β π‘πππ + (πβ π)π‘β π‘πππ
π + (π + πβ 1)π + π + (π β πβ 1)π
π + ππ + ππ β π + π + ππβ ππβ π
2π + 2ππβ 2π
2π + (πβ 1)2π
2[π + (πβ 1)π]
2 Γ ππ‘β π‘πππ
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 + (πβ 1)π
ππ = 58 + (πβ 1)2 β¦ β¦ β¦ β¦ . (π)
Now,
β2, 5, 12, β¦ β¦ β¦ ..
π = β2
π = 7
ππ = π + (πβ 1)π
ππ = β2 + (πβ 1)7 β¦ β¦ β¦ β¦ . . (ππ)
From (i) and (ii) we get,
58 + (πβ 1)2 = β2 + (πβ 1)7
58 + 2πβ 2 = β2 + 7πβ 7
56 + 2π = β9 + 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,
ππ be the first negative term
π = 105
π = 101 β 105
π = β4
ππ < 0
π + (πβ 1)π < 0
105 + (π β 1)(β4) < 0
105 + β4π + 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 + (πβ 1)π
994 = 105 + (π β 1)7
994 β 105 = (π β 1)7
889 = 7πβ 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 (πβ π), π, (π + π)
πβ π + π + π + π = 216
3π = 216
π = 72
According to question,
Product of the two smaller parts = 5040.
π(πβ π) = 5040
72(72β π) = 5040
72β π = 70
π = 2
πβ 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
β5
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 β5, 1, 11, β¦ β¦ β¦ .
π = β5
π = 1 β (β5)
π = 1 + 5
π = 6
π = 11 β 1
π = 10
Here, the common difference is not same
Hence, the nth term of an A.P canβt be 2n2 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 + (πβ 1)π
168 = 14 + (πβ 1)7
168 β 14 = (πβ 1)7
154 = 7πβ 7
154 = 7πβ 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 31st term of this A.P.
Solution:

It is given that,
π = 20
π7 = 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,
β12, β10, β8, β¦ . β¦ β¦ ,58
Reverse the term,
58, β¦ . , β8, β10, β12
π = 58
π = β2
ππ = π/2 [2π + (π β 1)π]
π8 = 8/2 [2(58) + (8 β 1)(β2)]
π8 = 4[116 + (7)(β2)]
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.