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 |
Other Exercises |
|
Related Study | NCERT Solutions for Class 10 Maths |
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 20th term of the AP = 9, 13, 17, 21, ....
(ii) the 35th term of AP 20, 17, 14, 11, .....
(iii) the 18th term of AP √2, √18, √50, √98 ....
(iv) the 9th term of the AP 3/4, 5/4, 7/4, 9/4, .....
(v) the 15th 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
nth term of the AP, an = a + (n – 1)d = 9 + (n – 1) ×4
∴ 20th term of the AP, a20 = 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
nth term of the AP, an = a + (n – 1)d
= 20 + (n – 1) × (-3)
∴ 35th 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
nth term of the AP, an = a + (n – 1)d = 3/4 + (n – 1) × (1/2)
∴ 9th term of AP, a9 = 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
nth term of the AP, an = a + (n – 1)d
= - 40 + (n – 1) × 25
∴ 15th term of the AP, a15 = - 40 + (15 – 1) × 25
= - 40 + 350
= 310
3. Find the 37thterm 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, T37 = (a + (37 – 1)d = a + 36d
= 6 + 36 × 7/4
= 6 + 63
= 69
∴ 37th term = 69
4. Find the 25th 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, T25 = a + (25 – 1)d = a + 24d
= 5 + 24 × (-1/2)
= 5 – 12
= - 7
∴ 25th 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
Tn = (4n – 10) [Given]
T1 = (4 × 1 – 10) = - 6
T2 = (4 × 2 – 10) = -2
T3 = (4 × 3 – 10) = 2
T4 = (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) T16 = 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, Tn = 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, Tn = 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,
an = - 47
⇒ 18 + (n – 1) × (-5/2) = - 47 [an = 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 nth term be 88
Then, Tn = 88
⇒ a + (n – 1)d = 88
⇒ 3 + (n – 1) × 5 = 88
⇒ 5n – 2 = 88
⇒ 5n = 90
⇒ n = 18
Hence, the 18th 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 nth term be 0.
Then, Tn = 0
⇒ a + (n – 1)d = 0
⇒ 72 + (n – 1) × (-4) = 0
⇒ 76 – 4n = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19th 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 nth term be 3.
Now, Tn = 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 14th 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 nth term of the given AP is -81. then,
an = - 81
⇒ 21 + (n – 1) × (-3) = - 81 [an = a + (n – 1)d]
⇒ -3(n – 1) = - 81 – 21 = -102
⇒ n – 1 = 102/3 = 34
⇒ n = 34 + 1 = 35
Hence, the 35th term of the given AP is – 81.
14. Which term of the AP 3, 8, 13, 18, ...... Will be 55 more than its 20th term?
Solution
Here, a = 3 and d = (8 – 3) = 5
The 20th term is given by
T20 = a + (20 – 1)d
= a + 19d
= 3 + 19 × 5 = 98
∴ Required term = (98 + 55) = 153
Let this be the nth term.
Then, Tn = 153
⇒ 3 + (n – 1) × 5 = 153
⇒ 5n = 155
⇒ n = 31
Hence, the 31st term will be 55 more than 20th term.
15. Which term of the AP 5, 15, 25, ........ will be 130 more than its 31st term?
Solution
Here, a = 5 and d = (15 – 5) = 10
The 31st term is given by
T31 = a + (31 - 1)d
= a + 30d
= 5 + 30 × 10
= 305
∴ Required term = (305 + 130) = 435
Let this be the nth term.
Then, Tn = 435
⇒ 5 + (n – 1) × 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44th term will be 130 more than its 31st term.
16. If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
Solution
In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n – 1)d
Now, we have:
T10 = a + (10 – 1)d
⇒ a + 9d = 52 ...(1)
T13 = a + (13 – 1)d = a + 12d ....(2)
T17 = a + (17 – 1)d = a + 16d ...(3)
But, it is given that T17 = 20 + T13
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,
an = 216
⇒ 6 + (n – 1) × 7 = 216 [an = 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)/2th term
= 16th 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,
an = - 62
⇒ 10 + (n – 1) × (-3) = - 62 [an = 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
= 13th 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,
an = 4.1/3
⇒ -(4/3) + (n – 1) × (1/3) = 13/3 [an = 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., 9th term and 10th term.
∴ Sum of the middle most terms of the given AP
= 9th term + 10th 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 8th term from the end of the AP 7, 10, 13, ........., 184.
Solution
Here, a = 7 and d = (10 – 7) = 3, l = 184 and n = 8th from the end.
Now, nth term from the end = [l – (n – 1)d]
8th term from the end = [184 – (8 – 1) × 3]
= [184 – (7 × 3)]
= (184 – 21)
= 163
Hence, the 8th term from the end is 163.
21. Find the 6th 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]
6th term from the end = [(-40) – (6 – 1) × (-3)]
= [-40 + (5 × 3)]
= (-40 + 15)
= - 25
Hence, the 6th 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,
an = 184
⇒ 3 + (n – 1) × 4 = 184 [an = 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,
an = - 150
⇒ 11 + (n – 1) × (-3) = -150 [an = 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,
an < 0
⇒ 121 + (n – 1) × (-4) < 0 [an = a + (n – 1)d]
⇒ 125 – 4n < 0
⇒ -4n < -125
⇒ n > 125/4 = 31.1/4
∴ n = 32
Hence, the 32nd 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,
an < 0
⇒ 20 + (n – 1) × (-3/4) < 0 [an = 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 28th term is the first negative term of the given AP.
26. The 7thterm of the an AP is -4 and its 13thterm is -16. Find the AP.
Solution
We have
T7 = a + (n – 1)d
⇒ a + 6d = - 4 ....(1)
T13 = 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 4thterm of an AP is zero. Prove that its 25thterm is triple its 11th term.
Solution
In the given AP, Let the first be a and the common difference be d.
Then, Tn = a + (n – 1)d
Now, T4 = a + (4 – 1)d
⇒ a + 3d = 0 ....(1)
⇒ a = - 3d
Again, T11 = a + (11 – 1)d = a + 10d
= - 3d + 10d = 7d [Using (1)]
Also, T25 = a + (11 – 1)d = a + 10d
= - 3d + 10d = 7d [Using (1)]
Also, T25 = a + (25 – 1)d = a + 24d
= -3d + 24d
= 21d [Using (1)]
i.e., T25 = 3 × 7d = (3 × T11)
Hence, 25th term is triple its 11th term.
28. The 8thterm of an AP is zero. Prove that its 38thterm is triple its 18th term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a8 = 0 [an = a + (n – 1)d]
⇒ a + (8 – 1)d = 0
⇒ a + 7d = 0
⇒ a = - 7d ....(1)
Now,
a38/a18 = {(a + (38 – 1)d}/{a + (18 – 1)d}
⇒ a38/a18 = (-7d + 37d)/(-7d + 17d) [From (1)]
⇒ a38/a18 = 30d/10d = 3
⇒ a38 = 3 × a18
Hence, the 38th term of the AP is triple its 18th term.
29. The 4th term of an AP is 11. The sum of the 5thand 7th 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,
a4 = 11
⇒ a + (4 – 1)d = 11 [an = a + (n – 1)d]
⇒ a + 3d = 11 ...(1)
Now,
a5 + a7 = 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 9th term of an AP. is – 32 and the sum of its 11thand 13th 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,
a9 = - 32
⇒ a + (9 – 1)d = - 32 [an = a + (n – 1)d]
⇒ a + 8d = -32 ...(1)
Now,
a11 + a13 = -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 7th term is -1 and 16th term is 17.
Solution
Let a be first term and d be the common difference of the AP. Then,
a7 = -1
⇒ a + (7 – 1)d = -1 [an = a + (n – 1)d]
⇒ a + 6d = -1 ...(1)
Also,
a16 = 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
∴ an = 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 4thterm of an AP is equal is 18 times its 18th term then find its 22nd term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
4 × a4 = 18 × a18 (Given)
⇒ 4(a + 3d) = 18(a + 17d) [an = 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
⇒ a22 = 0
Hence, the 22nd term of the AP is 0.
33. If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.
Solution
Let a be the first term and d be the common difference of the AP. Then,
10 × a10 = 15 × a15 (Given)
⇒ 10(a + 9d) = 15(a + 14d) [an = a + (n – 1)d]
⇒ 2(a + 9d) = 3(a + 14d)
⇒ 2a + 18d = 3a + 42d
⇒ a = - 24d
⇒ a + 24d = 0
⇒ a + (25 – 1)d = 0
⇒ a25 = 0
Hence, the 25th 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,
a1 + a2 + a3 + a4 = 1/2(a5 + a6 + a7 + a8) (Given)
⇒ a + (a + d) + (a + 2d) + (a + 3d) = 1/2[(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)]
[an = 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 2nd and 7th terms of AP is 30. If its 15th term is 1 less than twice its 8th term, find AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a2 + a7 = 30 (Given)
∴ (a + d) + (a + 6d) = 30 [an = a + (n – 1)d]
⇒ 2a + 7d = 30 ....(1)
Also,
a15 = 2a8 – 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, a2 = a + d = 1 + 4 = 5
a3 = 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 tn and Tn, 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
tn = 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
Tn = A + (n – 1)D
⇒ 3 + (n – 1) × 7
⇒ 7n – 4
Now, tn = Tn
⇒ 61 + 2n = 7n – 4
⇒ 65 = 5n
⇒ n = 13
Hence, the 13 terms of the A1’s are the same.
37. The 17th term of AP is 5 more than twice its 8th term. If the 11th 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,
a17 = 2a8 + 5 (Given)
∴ a + 16d = 2(a + 7d) + 5 [an = a + (n – 1)d]
⇒ a + 16d = 2a + 14d + 5
⇒ a – 2d = - 5 ...(1)
Also,
a11 = 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
∴ an = a + (n – 1)d
= 3 + (n – 1) × 4
= 4n – 1
Hence, the nth term of AP is (4n – 1).
38. The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
Solution
Let a be the first term and d be common difference of the AP. Then,
a24 = 2a10 (Given)
⇒ a + 23d = 2(a + 9d) [an = a + (n – 1)d]
⇒ a + 23d = 2a + 18d
⇒ 2a – a = 23d – 18d
⇒ a = 5d ...(1)
Now,
a72/a15 = (a + 71d)/(a + 14d)
⇒ a72/a15 = (5d + 71d)/(5d + 14d) [From (1)]
⇒ a72/a15 = 76d/19d = 4
⇒ a22 = 4 × a15
Hence, the 72nd term of the AP is 4 times its 15th term.
39. The 19th term of an AP is equal to 3 times its 6th term. if its 9th term is 19, find the AP.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a19 = 3a6 (Given)
⇒ a + 18d = 3(a + 5d) [an = a + (n – 1)d]
⇒ a + 18d = 3a + 15d
⇒ 3a – a = 18d – 15d
⇒ 2a = 3d ...(1)
Also,
a9 = 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,
a2 = a + d = 3 + 2 = 5
a3 = 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, Tn = a + (n – 1)d
⇒ Tp = a + (p + q – 1)d ...(i)
⇒ Tq = 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, Tp+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
Tn = a + (n – 1)d ...(1)
Similarly, nth term from the end is given by
Tn = {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,
an = 96
⇒ 12 + (n – 1) × 6 = 96 [an = a + (n – 1)d]
⇒ 6n + 6 = 96
⇒ 6n = 96 – 6 = 90
⇒ n = 15
Hence, these are 15 two-digit numbers divisible by 6.
43. How many two-digits numbers are divisible by 3?
Solution
The two-digit 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,
an = 99
⇒ 12 + (n – 1) × 3 = 99 [an = a + (n – 1)d]
⇒ 3n + 9 = 99
⇒ 3n = 99 – 9 = 90
⇒ n = 30
Hence, there are 30 two-digit numbers divisible by 3.
44. How many three-digit numbers are divisible by 9?
Solution
The three-digit 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,
an = 999
⇒ 108 + (n – 1) × 9 = 999 [an = a + (n – 1)d]
⇒ 9n + 99 = 999
⇒ 9n = 999 – 99 = 900
⇒ n = 100
Hence, there are 100 three-digit 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,
an = 990
⇒ 110 + (n – 1) × 10 = 990 [an = 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:
Tn = a + (n – 1)d
⇒ 11 = 43 + (n – 1)(-2)
⇒ 11 = 45 – 2n
⇒ 34 = 2n
⇒ n = 17
Hence, the 17th 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.