RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11A Class 10 Maths
Chapter Name  RS Aggarwal Chapter 11 Arithmetic Progression 
Book Name  RS Aggarwal Mathematics for Class 10 
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Exercise 11A Solutions
1. Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.
(i) 9, 15, 21, 27, …
(ii) 11, 6, 1, – 4, ....
(iii) 1, 5/6, 2/3, 1/2, .....
(iv) √2, √8, √18, √32 ....
(v) √20, √45, √80, √125 ....
Solution
(i) The given progression 9, 15, 21, 27, …..
Clearly, 15 – 9 = 21 – 15 = 27 – 21 = 6 (Constant)
Thus, each term differs from its preceding term by 6. So, the given progression is an AP.
First term = 9
Common difference = 6
Next term of the AP = 27 + 6 = 33
(ii) The given progression 11, 6, 1, – 4, …….
Clearly, 6 – 11 = 1 – 6 = –4 – 1 = –5 (Constant)
Thus, each term differs from its preceding term by 6. So, the given progression is an AP.
First term = 11
Common difference = –5
Next term of the AP =  4 + (5) = 9
(iii) The given progression – 1, 5/6, 2/3, 1/2, ....
Clearly, 5/6 – (1) = 2/3 – (5/6)
= 1/2 – (2/3) = 1/6 (constant)
Thus, each term differs from its preceding term by 1/6 So, the given progression is an AP.
First term = –1
Common difference = 1/6
Next tern of the AP = 1/2 + 1/6 = 2/6 = 1/3
(iv) The given progression √2, √8, √18, √32 ....
This sequence can be written as √2, 2√2, 3√2, 4√2, ......
Clearly, 2√2  √2 = 3√2  2√2 = 4√2  3√2 = 2 (Constant)
Thus, each term differs from its preceding term by √2. So, the given progression is an AP.
First term = √2
Common difference = √2
Next term of the AP = 4√2 + √2 = 5√2 = √50
(v)
2. Find:
(i) the 20^{th} term of the AP = 9, 13, 17, 21, ....
(ii) the 35^{th} term of AP 20, 17, 14, 11, .....
(iii) the 18^{th} term of AP √2, √18, √50, √98 ....
(iv) the 9^{th} term of the AP 3/4, 5/4, 7/4, 9/4, .....
(v) the 15^{th} term of the AP 40,  15, , 10, 35, ....
Solution
(i) The given AP is 9, 13, 17, 21, .....
First term, a = 9
Common difference, d = 13 – 9 = 4
n^{th} term of the AP, a_{n} = a + (n – 1)d = 9 + (n – 1) ×4
∴ 20^{th} term of the AP, a_{20} = 9 + (20 – 1) ×4 = 9 + 76 = 85
(ii) The given AP is 20, 17, 14, 11, .......
First term, a = 20
Common difference, d = 17 – 20 = 3
n^{th} term of the AP, a_{n} = a + (n – 1)d
= 20 + (n – 1) × (3)
∴ 35^{th} term of the AP, a35 = 20 + (35 – 1) × (3)
= 20 – 102
=  82
(iii)
(iv) The given AP is 3/4, 5/4, 7/4, 9/4, ......
First term, a = 3/4
Common difference, d = 5/4 – 3/4 = 2/4 = 1/2
n^{th} term of the AP, a_{n} = a + (n – 1)d = 3/4 + (n – 1) × (1/2)
∴ 9^{th} term of AP, a_{9} = 3/4 + (9 – 1) × 1/2
= 3/4 + 4
= 19/4
(v) The given AP is 40, 15, 10, 35, .......
First term, a =  40
Common difference, d =  15 – (40) = 25
n^{th} term of the AP, a_{n} = a + (n – 1)d
=  40 + (n – 1) × 25
∴ 15^{th} term of the AP, a_{15} =  40 + (15 – 1) × 25
=  40 + 350
= 310
3. Find the 37^{th}term of the AP, 6, 7.3/4, 9.1/2, 11.1/4, ......
Solution
The given AP is 6, 7.3/4, 9.1/2, 11.1/4, ......
First term, a = 6 and common difference, d = 7.3/4 – 6
⇒ 31/4 – 6
= (31 – 24)/4
= 7/4
Now, T_{37} = (a + (37 – 1)d = a + 36d
= 6 + 36 × 7/4
= 6 + 63
= 69
∴ 37^{th} term = 69
4. Find the 25^{th} term of the AP 5, 4.1/2, 4, 3.1/2, 3, …
Solution
The given AP is 5, 4.1/2, 4, 3.1/2, 3, …
First term = 5
Common difference = 4.1/2 – 5 ⇒ 9/2 – 5
⇒ (9 – 10)/2
= 1/2
∴ a = 5 and d = (1/2)
Now, T_{25} = a + (25 – 1)d = a + 24d
= 5 + 24 × (1/2)
= 5 – 12
=  7
∴ 25^{th} term = 7
5. Find the nth term of each of the following Aps:
(i) 5, 11, 17, 23, ....
(ii) 16, 9, 2, 5, ........
Solution
(i) (6n – 1)
(ii) (23 – 7n)
6. If the nth term of a progression is (4n – 10) show that it is AP. Find its
(i) First term,
(ii) common difference
(iii) 16 the term.
Solution
T_{n} = (4_{n }– 10) [Given]
T_{1} = (4 × 1 – 10) =  6
T_{2} = (4 × 2 – 10) = 2
T_{3} = (4 × 3 – 10) = 2
T_{4} = (4 × 4 – 10) = 6
Clearly, [ 2 – (6)]
= [2 – (2)]
= [6 – 2]
= 4 (Constant)
So, the terms 6, 2, 2, 6, ...... forms an AP.
Thus, we have
(i) First term = 6
(ii) Common difference = 4
(iii) T_{16} = a + (n – 1)d
= a + 15d
= 6 + 15 × 4
= 54
7. How many terms are there in the AP 6, 1, 0, 14, 18, ......, 174?
Solution
In the given AP, a = 6 and d = (10 – 6) = 4
Suppose that there are n terms in the given AP.
Then, T_{n} = 174
⇒ a + (n – 1)d = 174
⇒ 6 + (n – 1) × 4 = 174
⇒ 2 + 4n = 174
⇒ 4n = 172
⇒ n = 43
Hence, there are 43 terms in the given AP.
8. How many terms are there in the AP 41, 38, 35, ........, 8?
Solution
In the given AP, a = 41 and d = (38 – 41) =  3
Suppose that there are n terms in the given AP.
Then, T_{n} = 8
⇒ a + (n – 1)d = 8
⇒ 41 + (n – 1) × (3) = 8
⇒ 44 – 3n = 8
⇒ 3n = 36
⇒ n = 12
Hence, there are 12 terms in the given AP.
9. How many terms are there in AP 18, 15.1/2, 13, ........, 47?
Solution
The given AP is 18, 15.1/2, 13, ......, 47
First term, a = 18
Common difference, d = 15.1/2 – 18
= 31/2 – 18
= (31 – 36)/2
= (5/2)
Suppose there are n terms in the given AP. Then,
a_{n} =  47
⇒ 18 + (n – 1) × (5/2) =  47 [a_{n} = a + (n – 1)d]
⇒  5/2(n – 1) =  47 – 18 =  65
⇒ n – 1 =  65 × (2/5) = 26
⇒ n = 26 + 1 = 27
Hence, there are 27 terms in the given AP.
10. Which term of the AP 3, 8, 13, 18, ...... is 88?
Solution
In the given AP, first term, a = 3 and common difference, a = (8 – 3) = 5.
Let’s its n^{th} term be 88
Then, T_{n} = 88
⇒ a + (n – 1)d = 88
⇒ 3 + (n – 1) × 5 = 88
⇒ 5n – 2 = 88
⇒ 5n = 90
⇒ n = 18
Hence, the 18^{th} term of the given AP is 88.
11. Which term of AP 72, 68, 64, 60, ........ is 0?
Solution
In the given AP, first term, a = 72 and common difference, d = (68 – 72)
= 4
Let its n^{th} term be 0.
Then, T_{n} = 0
⇒ a + (n – 1)d = 0
⇒ 72 + (n – 1) × (4) = 0
⇒ 76 – 4n = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19^{th} term of the given AP is 0.
12. Which term of the AP 5/6, 1, 1.1/6 , 1.1/3, ....... is 3?
Solution
In the given AP, first term = 5/6 and common difference, d = (1 – 5/6 = 1/6)
Let its n^{th} term be 3.
Now, T_{n} = 3
⇒ a + (n – 1)d = 3
⇒ 5/6 + (n – 1) × 1/6 = 3
⇒ 2/3 + n/6 = 3
⇒ n/6 = 7/3
⇒ n = 14
Hence, the 14^{th} term of the given AP is 3.
13. Which term of the AP 21, 18, 15, .......... is 81?
Solution
The given AP is 21, 18, 15, .....
First term, a = 21
Common difference, d = 18 – 21 =  3.
Suppose n^{th} term of the given AP is 81. then,
a_{n} =  81
⇒ 21 + (n – 1) × (3) =  81 [a_{n} = a + (n – 1)d]
⇒ 3(n – 1) =  81 – 21 = 102
⇒ n – 1 = 102/3 = 34
⇒ n = 34 + 1 = 35
Hence, the 35^{th} term of the given AP is – 81.
14. Which term of the AP 3, 8, 13, 18, ...... Will be 55 more than its 20^{th} term?
Solution
Here, a = 3 and d = (8 – 3) = 5
The 20^{th} term is given by
T_{20} = a + (20 – 1)d
= a + 19d
= 3 + 19 × 5 = 98
∴ Required term = (98 + 55) = 153
Let this be the n^{th} term.
Then, T_{n }= 153
⇒ 3 + (n – 1) × 5 = 153
⇒ 5n = 155
⇒ n = 31
Hence, the 31^{st} term will be 55 more than 20^{th} term.
15. Which term of the AP 5, 15, 25, ........ will be 130 more than its 31^{st} term?
Solution
Here, a = 5 and d = (15 – 5) = 10
The 31^{st} term is given by
T_{31} = a + (31  1)d
= a + 30d
= 5 + 30 × 10
= 305
∴ Required term = (305 + 130) = 435
Let this be the n^{th} term.
Then, T_{n} = 435
⇒ 5 + (n – 1) × 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44^{th} term will be 130 more than its 31^{st} term.
16. If the 10^{th} term of an AP is 52 and 17^{th} term is 20 more than its 13^{th} term, find the AP
Solution
In the given AP, let the first term be a and the common difference be d.
Then, T_{n} = a + (n – 1)d
Now, we have:
T_{10} = a + (10 – 1)d
⇒ a + 9d = 52 ...(1)
T_{13} = a + (13 – 1)d = a + 12d ....(2)
T_{17} = a + (17 – 1)d = a + 16d ...(3)
But, it is given that T_{17} = 20 + T_{13 }
i.e., a + 16d = 20 + a + 12d
⇒ 4d = 20
⇒ d = 5
On substituting d = 5 in (1), we get:
a + 9 × 5 = 52
⇒ a = 7
Thus, a = 7 and a = 5
∴ The terms of the AP are 7, 12, 17, 22, ......
17. Find the middle term of the AP 6, 13, 20, ........., 216.
Solution
The given AP is 6, 13, 20, ......, 216.
First term, a = 6
Common difference, d = 13 – 6 = 7
Suppose these are n terms in the given AP. Then,
a_{n} = 216
⇒ 6 + (n – 1) × 7 = 216 [a_{n} = a + (n – 1)d]
⇒ 7(n – 1) = 216 – 6 = 210
⇒ n – 1 = 210/7 = 30
⇒ n = 30 + 1 = 31
Thus, the given AP contains 31 terms,
∴ Middle term of the given AP
= (31 + 1)/2^{th} term
= 16^{th} term
= 6 + (16 – 1) × 7
= 6 + 105
= 111
Hence, the middle term of the given AP is 111.
18. Find the middle term of the AP 10, 7, 4, ......, 62.
Solution
The given AP is 10, 7, 4, ......., 62.
First term, a = 10
Common difference, d = 7 – 10 =  3
Suppose these are n terms in the given AP. Then,
a_{n} =  62
⇒ 10 + (n – 1) × (3) =  62 [a_{n} = a + (n – 1)d]
⇒ 3(n – 1) =  62 – 10 =  72
⇒ n – 1 = 72/3 = 24
⇒ n = 24 + 1 = 25
Thus, the given AP contains 25 terms.
∴ Middle term of the given AP
= (25 + 1)/2th term
= 13^{th} term
= 10 + (13 – 1) × (3)
= 10 – 36
=  26
Hence, the middle term of the given AP is – 26.
19. Find the sum of two middle most terms of the AP –(4/3),  1, 2/3, ......., 4.1/3.
Solution
The given AP is –(4/3), 1, 2/3, .........4.1/3.
First term, a = (4/3)
Common difference, d =  1 – (4/3)
= 1 + 4/3
= 1/3
Suppose there are n terms in the given AP. Then,
a_{n} = 4.1/3
⇒ (4/3) + (n – 1) × (1/3) = 13/3 [a_{n} = a + (n – 1)d]
⇒ 1/3(n – 1) = 13/3 + 4/3 = 17/3
⇒ n – 1 = 17
⇒ n = 17 + 1 = 18
Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.
The middle terms of the given AP are (18/2)^{th} terms and (18/2 + 1)^{th} terms i.e., 9^{th} term and 10^{th} term.
∴ Sum of the middle most terms of the given AP
= 9^{th} term + 10^{th} term
= [4/3 + (9 – 1) × 1/3] + [4/3 + (10 – 1) × 1/3]
= 4/3 + 8/3 – 4/3 + 3
= 3
Hence, the sum of the middle most terms of the given AP is 3.
20. Find the 8^{th} term from the end of the AP 7, 10, 13, ........., 184.
Solution
Here, a = 7 and d = (10 – 7) = 3, l = 184 and n = 8^{th} from the end.
Now, n^{th} term from the end = [l – (n – 1)d]
8^{th} term from the end = [184 – (8 – 1) × 3]
= [184 – (7 × 3)]
= (184 – 21)
= 163
Hence, the 8^{th} term from the end is 163.
21. Find the 6^{th} term from the end of the AP 17, 14, 11, .........., (40).
Solution
Here, a = 7 and d = (14 – 17) =  3, l = (40) and n = 6
Now, nth term from the end = [l – (n – 1)d]
6^{th} term from the end = [(40) – (6 – 1) × (3)]
= [40 + (5 × 3)]
= (40 + 15)
=  25
Hence, the 6^{th} term from the end is – 25.
22. Is 184 a term of AP 3, 7, 11, 15, ........?
Solution
The given AP is 3, 7, 11, 15, ......
Here, a = 3 and d = 7 – 3 = 4
Let the nth term of the given AP be 184. Then,
a_{n} = 184
⇒ 3 + (n – 1) × 4 = 184 [a_{n} = a + (n – 1)d]
⇒ 4n – 1 = 184
⇒ 4n = 185
⇒ n = 185/4 = 46.1/4
But, the number of terms cannot be a fraction.
Hence, 184, is not a term of the given AP.
23. Is 150 a term of the AP 11, 8, 5, 2, ........?
Solution
The given AP is 11, 8, 5, 2, ......
Here, a = 11 and d = 8 – 11 = 3
Let the nth term of the given AP be 150. Then,
a_{n} =  150
⇒ 11 + (n – 1) × (3) = 150 [a_{n} = a + (n – 1)d]
⇒  3n + 14 =  150
⇒  3n =  164
⇒ n = 164/3 = 54.2/3
But, the number of terms cannot be a fraction.
Hence, 150 is not a term of the given AP.
24. Which term of the AP 121, 117, 113, ........ is its first negative term?
Solution
The given AP is 121, 117, 113, ......
Here, a = 121 and d = 117 – 121 =  4
Let the nth term of the given AP be the first negative term. Then,
a_{n} < 0
⇒ 121 + (n – 1) × (4) < 0 [a_{n} = a + (n – 1)d]
⇒ 125 – 4n < 0
⇒ 4n < 125
⇒ n > 125/4 = 31.1/4
∴ n = 32
Hence, the 32^{nd} term is the first negative term of the given AP.
25. Which term of AP 20, 19.1/4, 18.1/2, 17.3/4, ........ is the first negative term?
Solution
The given AP is 20, 19.1/4, 18.1/2, 17.3/4, ....
Here, a = 20 and d = 19.1/4 – 20
= 77/4 – 20
= (77  80)/4
= (3/4)
Let the nth term of the given AP be the first negative term, Then,
a_{n} < 0
⇒ 20 + (n – 1) × (3/4) < 0 [a_{n} = a + (n – 1)d]
⇒ 20 + 3/4 – 3/4.n < 0
⇒ 83/4 – 3/4.n < 0
⇒ 3/4.n <  83/4
⇒ n > 83/3 = 27.2/3
∴ n = 28
Hence, the 28^{th} term is the first negative term of the given AP.
26. The 7^{th}term of the an AP is 4 and its 13^{th}term is 16. Find the AP.
Solution
We have
T_{7 }= a + (n – 1)d
⇒ a + 6d =  4 ....(1)
T_{13} = a + (n 1)d
⇒ a + 12d =  16 ...(2)
On solving (1) and (2), we get
a = 8 and d =  2
Thus, first term = 8 and common difference = 2
∴ The term of the AP are 8, 6, 4, 2, .......
27. The 4^{th}term of an AP is zero. Prove that its 25^{th}term is triple its 11^{th} term.
Solution
In the given AP, Let the first be a and the common difference be d.
Then, T_{n }= a + (n – 1)d
Now, T_{4} = a + (4 – 1)d
⇒ a + 3d = 0 ....(1)
⇒ a =  3d
Again, T_{11 }= a + (11 – 1)d = a + 10d
=  3d + 10d = 7d [Using (1)]
Also, T_{25} = a + (11 – 1)d = a + 10d
=  3d + 10d = 7d [Using (1)]
Also, T_{25} = a + (25 – 1)d = a + 24d
= 3d + 24d
= 21d [Using (1)]
i.e., T_{25} = 3 × 7d = (3 × T_{11})
Hence, 25^{th} term is triple its 11^{th} term.
28. The 8^{th}term of an AP is zero. Prove that its 38^{th}term is triple its 18^{th} term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{8 }= 0 [a_{n} = a + (n – 1)d]
⇒ a + (8 – 1)d = 0
⇒ a + 7d = 0
⇒ a =  7d ....(1)
Now,
a_{38}/a_{18} = {(a + (38 – 1)d}/{a + (18 – 1)d}
⇒ a_{38}/a_{18} = (7d + 37d)/(7d + 17d) [From (1)]
⇒ a_{38}/a_{18} = 30d/10d = 3
⇒ a_{38 }= 3 × a_{18}
Hence, the 38^{th} term of the AP is triple its 18^{th} term.
29. The 4^{th }term of an AP is 11. The sum of the 5^{th}and 7^{th} terms of this AP is 34. Find its common difference.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{4 }= 11
⇒ a + (4 – 1)d = 11 [a_{n }= a + (n – 1)d]
⇒ a + 3d = 11 ...(1)
Now,
a_{5} + a_{7} = 34 (Given)
⇒ (a + 4d) + (a + 6d) = 34
⇒ 2a + 10d = 34
⇒ a + 5d = 17 ...(2)
From (1) and (2), we get
11 – 3d + 5d = 17
⇒ 2d = 17 – 11 = 6
⇒ d = 3
Hence, the common difference of the AP is 3.
30. The 9^{th }term of an AP. is – 32 and the sum of its 11^{th}and 13^{th} terms is – 94. Find the common difference of the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{9} =  32
⇒ a + (9 – 1)d =  32 [a_{n} = a + (n – 1)d]
⇒ a + 8d = 32 ...(1)
Now,
a_{11} + a_{13} = 94 (Given)
⇒ (a + 10d) + (a + 12d) =  94
⇒ 2a + 22d =  94
⇒ a + 11d = 47 ...(2)
From (1) and (2), we get
32 – 8d + 11d =  47
⇒ 3d =  47 + 32 =  15
⇒ d = 5
Hence, the common difference of the AP is 5.
31. Determine the nth term of the AP whose 7^{th }term is 1 and 16^{th }term is 17.
Solution
Let a be first term and d be the common difference of the AP. Then,
a_{7} = 1
⇒ a + (7 – 1)d = 1 [a_{n} = a + (n – 1)d]
⇒ a + 6d = 1 ...(1)
Also,
a_{16} = 17
⇒ a + 15d = 17 ...(2)
From (1) and (2), we get
1  6d + 15d = 17
⇒ 9d = 17 + 1 = 18
⇒ d = 2
Putting d = 2 in (1), we get
a + 6 × 2 = 1
⇒ a =  1 – 12 = 13
∴ a_{n} = a + (n – 1)d
=  13 + (n – 1) × 2
= 2n – 15
Hence, the nth term of the AP is (2n – 15).
32. If 4 times the 4^{th}term of an AP is equal is 18 times its 18^{th} term then find its 22^{nd} term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
4 × a_{4} = 18 × a_{18} (Given)
⇒ 4(a + 3d) = 18(a + 17d) [a_{n} = a + (n – 1)d]
⇒ 2(a + 3d) = 9(a + 17d)
⇒ 2a + 6d = 9a + 153d
⇒ 7a =  147d
⇒ a =  21d
⇒ a + 21d = 0
⇒ a + (22 – 1)d = 0
⇒ a_{22} = 0
Hence, the 22^{nd} term of the AP is 0.
33. If 10 times the 10^{th} term of an AP is equal to 15 times the 15^{th} term, show that its 25^{th} term is zero.
Solution
Let a be the first term and d be the common difference of the AP. Then,
10 × a_{10} = 15 × a_{15} (Given)
⇒ 10(a + 9d) = 15(a + 14d) [a_{n} = a + (n – 1)d]
⇒ 2(a + 9d) = 3(a + 14d)
⇒ 2a + 18d = 3a + 42d
⇒ a =  24d
⇒ a + 24d = 0
⇒ a + (25 – 1)d = 0
⇒ a_{25} = 0
Hence, the 25^{th} term of the AP is 0.
34. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Solution
Let the common difference of the AP be d.
First term, a = 5
Now,
a_{1} + a_{2} + a_{3} + a_{4} = 1/2(a_{5} + a_{6} + a_{7} + a_{8}) (Given)
⇒ a + (a + d) + (a + 2d) + (a + 3d) = 1/2[(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)]
[a_{n} = a + (n – 1)d]
⇒ 4a + 6d = ½(4a + 22d)
⇒ 8a + 12d = 4a + 22d
⇒ 22d – 12d = 8a – 4a
⇒ 10d = 4a
⇒ d = (2/5)a
⇒ d = 2/5 × 5 = 2 (a = 5)
Hence, the common difference of the AP is 2.
35. The sum of the 2^{nd} and 7^{th} terms of AP is 30. If its 15^{th} term is 1 less than twice its 8^{th} term, find AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{2} + a_{7} = 30 (Given)
∴ (a + d) + (a + 6d) = 30 [a_{n} = a + (n – 1)d]
⇒ 2a + 7d = 30 ....(1)
Also,
a_{15 }= 2a_{8} – 1 (Given)
⇒ a + 14d = 2(a + 7d) – 1
⇒ a + 14d = 2a + 14d – 1
⇒ a =  1
⇒ a = 1
Putting a = 1 in (1), we get
2 × 1 + 7d = 30
⇒ 7d = 30 – 2 = 28
⇒ d = 4
So, a_{2 }= a + d = 1 + 4 = 5
a_{3} = a + 2d = 1 + 2 × 4 = 9
Hence, the AP is 1, 5, 9, 13, .......
36. For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ..... and 3, 10, 17, ..... are equal?
Solution
Let the term of the given progressions be t_{n} and T_{n}, respectively.
The first AP is 63, 65, 67, ......
Let its first term be a and common difference be d.
Then a = 63 and d = (65 – 63) = 2
So, its nth term is given by
t_{n} = a + (n – 1)d
⇒ 63 + (n – 1) × 2
⇒ 61 + 2n
The second AP is 3, 10, 17, .....
Let its first term be A and common difference be D.
Then A = 3 and D = (10 – 3) = 7
So, its nth term is given by
T_{n} = A + (n – 1)D
⇒ 3 + (n – 1) × 7
⇒ 7n – 4
Now, t_{n} = T_{n }
⇒ 61 + 2n = 7n – 4
⇒ 65 = 5n
⇒ n = 13
Hence, the 13 terms of the A1’s are the same.
37. The 17^{th} term of AP is 5 more than twice its 8^{th} term. If the 11^{th} term of the AP is 43, find its nth term.
Solution
Let a be the first term and d be common difference of the AP. Then,
a_{17} = 2a_{8} + 5 (Given)
∴ a + 16d = 2(a + 7d) + 5 [a_{n} = a + (n – 1)d]
⇒ a + 16d = 2a + 14d + 5
⇒ a – 2d =  5 ...(1)
Also,
a_{11} = 43 (Given)
⇒ a + 10d = 43 ...(2)
From (1) and (2), we get
5 + 2d + 10d = 43
⇒ 12d = 43 + 5 = 48
⇒ d = 4
Putting d = 4 in (1), we get
a – 2 × 4 =  5
⇒ a =  5 + 8 = 3
∴ a_{n} = a + (n – 1)d
= 3 + (n – 1) × 4
= 4n – 1
Hence, the nth term of AP is (4n – 1).
38. The 24^{th} term of an AP is twice its 10^{th} term. Show that its 72^{nd} term is 4 times its 15^{th} term.
Solution
Let a be the first term and d be common difference of the AP. Then,
a_{24} = 2a_{10} (Given)
⇒ a + 23d = 2(a + 9d) [a_{n} = a + (n – 1)d]
⇒ a + 23d = 2a + 18d
⇒ 2a – a = 23d – 18d
⇒ a = 5d ...(1)
Now,
a_{72}/a_{15} = (a + 71d)/(a + 14d)
⇒ a_{72}/a_{15} = (5d + 71d)/(5d + 14d) [From (1)]
⇒ a_{72}/a_{15} = 76d/19d = 4
⇒ a_{22 }= 4 × a_{15}
Hence, the 72^{nd} term of the AP is 4 times its 15^{th} term.
39. The 19^{th} term of an AP is equal to 3 times its 6^{th} term. if its 9^{th} term is 19, find the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{19} = 3a_{6} (Given)
⇒ a + 18d = 3(a + 5d) [a_{n} = a + (n – 1)d]
⇒ a + 18d = 3a + 15d
⇒ 3a – a = 18d – 15d
⇒ 2a = 3d ...(1)
Also,
a_{9 }= 19 (Given)
⇒ a + 8d = 10 (2)
From (1) and (2), we get
3d/2 + 8d = 19
⇒ (3d + 16d)/2 = 19
⇒ 19d = 38
⇒ d = 2
Putting d = 2in (1), we get:
2a = 3 × 2 = 6
⇒ a = 3
So,
a_{2} = a + d = 3 + 2 = 5
a_{3} = a + 2d = 3 + 2 × 2 = 7, .....
Hence, the AP is 3, 5, 7, 9, .....
40. If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.
Solution
In the given given AP, let the first be a and the common difference be d.
Then, T_{n} = a + (n – 1)d
⇒ T_{p} = a + (p + q – 1)d ...(i)
⇒ T_{q} = a + (q – 1)d = p ....(ii)
On subtracting (i) from (ii), we get:
(q – p)d = (p – q)
⇒ d =  1
Putting d =  1 in (i), we get:
a = (p + q – 1)
Thus, a = (p + q – 1) and d = 1
Now, T_{p+q} = a + (p + q – 1)d
= (p + q – 1) + (p + q – 1)(1)
= (p + q – 1) – (p + q – 1) = 0
Hence, the (p + q)^{th} term is 0 (zero).
41. The first and last terms of an AP are a and 1 respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a + 1).
Solution
In the given AP, first term = a and last term = l.
Let the common difference be d.
Then, nth term from the beginning is given by
T_{n} = a + (n – 1)d ...(1)
Similarly, nth term from the end is given by
T_{n} = {l – (n – 1)d} ...(2)
Adding (1) and (2), we get
a + (n – 1)d + {l + (n – 1)d}
= a + (n – 1)d + l – (n – 1)d
= a + 1
Hence, the sum of the nth term from the beginning and the nth term from the end (a + 1).
42. How many two digit number are divisible by 6?
Solution
The two digit numbers divisible by 6 are 12, 18, 24, ...., 96
Clearly, these number are in AP.
Here, a = 12 and d = 18 – 12 = 6
Let this AP contains n terms, Then,
a_{n} = 96
⇒ 12 + (n – 1) × 6 = 96 [a_{n} = a + (n – 1)d]
⇒ 6n + 6 = 96
⇒ 6n = 96 – 6 = 90
⇒ n = 15
Hence, these are 15 twodigit numbers divisible by 6.
43. How many twodigits numbers are divisible by 3?
Solution
The twodigit numbers divisible by 3 are 12, 15, 18, ...., 99
Clearly, these number are in AP.
Here, a = 12 and d = 15 – 12 = 3
Let this AP contains n terms. Then,
a_{n} = 99
⇒ 12 + (n – 1) × 3 = 99 [a_{n} = a + (n – 1)d]
⇒ 3n + 9 = 99
⇒ 3n = 99 – 9 = 90
⇒ n = 30
Hence, there are 30 twodigit numbers divisible by 3.
44. How many threedigit numbers are divisible by 9?
Solution
The threedigit numbers divisible by 9 are 108, 117, 126, .... 999.
Clearly, these number are in AP.
Here, a = 108 and d = 117 – 108 = 9
Let this AP contains n terms. Then,
a_{n} = 999
⇒ 108 + (n – 1) × 9 = 999 [a_{n} = a + (n – 1)d]
⇒ 9n + 99 = 999
⇒ 9n = 999 – 99 = 900
⇒ n = 100
Hence, there are 100 threedigit numbers divisible by 9.
45. How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
Solution
The numbers which are divisible by both 2 and 5 are divisible by 10 also.
Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, ...990.
Clearly, these number are in AP.
Here, a = 110 and d = 120 – 110 = 10
Let this AP contains n terms. Then,
a_{n} = 990
⇒ 110 + (n – 1) × 10 = 990 [a_{n} = a + (n – 1)d]
⇒ 10n + 100 = 990
⇒ 10n = 990 – 100 = 890
⇒ n = 89
Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.
46. In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?
Solution
The numbers of rose plants in consecutive rows are 43, 41, 39, ..., 11.
Difference of rose plants between two consecutive rows = (41 – 43) = (39 – 41) = 2
[Constant]
So, the given progression is an AP
Here, first term = 43
Common difference =  2
Last term 11
Let b be the last term, then, we have:
T_{n} = a + (n – 1)d
⇒ 11 = 43 + (n – 1)(2)
⇒ 11 = 45 – 2n
⇒ 34 = 2n
⇒ n = 17
Hence, the 17^{th} term is 11 or there are 17 rows in the flower bed.
47. A sum of₹ 2800 is to be used to award four prizes. If each prize after the first is ₹ 200 less than the preceding prize, find the value of each of the prizes.
Solution
Let the amount of the first prize be ₹ a
Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.
Amount of the second prize = ₹ (a – 200)
Amount of the third prize = ₹ (a – 2 × 200) = (a – 400)
Amount of the fourth prize = ₹ (a – 3 × 200) = (a – 600)
Now,
Total sum of the four prizes = 2,800
∴ ₹ a + ₹(a – 200) + ₹(a – 400) + ₹(a – 600)
= ₹ 2,800
⇒ 4a – 1200 = 2800
⇒ 4a = 2800 + 1200 = 4000
⇒ a = 1000
∴ Amount of the first prize = ₹ 1,000
Amount of the second prize = ₹(1000 – 200) = ₹ 800
Amount of the third prize = ₹(1000 – 400) = ₹ 600
Amount of the fourth prize = ₹(1000 – 600) = ₹ 400
Hence, the value of each of the prizes is ₹ 1,000, ₹ 800, ₹ 600 and ₹ 400.