RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions
Exercise 11A
Question 1:
The given progression is 3, 9, 15, 21 …..
Clearly (9 – 3) = (15 – 9) = (21 – 15) = 6 which is constant
Thus, each term differs from its preceding term by 6
So, the given progression is an AP
Its first term = 3 and the common difference = 6
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Question 2:
The given progression is 16, 11, 6, 1, -4 ….
Clearly (11 – 16) = (1 – 6) = (-4 – 1) = – 5 which is constant
Thus, each term differs from its preceding term by – 5
So the given progression is an AP
Its first term = 16 and the common difference = – 5
Question 3:
(i) The given AP is 1, 5, 9, 13, 17…..
Its first term = 1 and common difference = (5 – 1) = 4
∴ a = 1 and d = 4
The nth term of the AP is given by
Tn = a + (n-1) d
T20 = 1 + (20-1) x 4 = 1+ 76 = 77
Hence, the 20th term is 77
(ii) The given AP is 6, 9, 12, 15 ……
Its first term = 6 and common difference = (9 – 6) = 3
∴ a = 6, d = 3
The nth term of the AP is given by
Tn = a + (n-1) d
T35 = 6 + (35-1) x 3 = 6+ 102 = 108
Hence, the 35th term is 108
(iii) The given AP is 5, 11, 17, 23 …..
Its first term = 5, and common difference = (11 – 5) = 6
∴ a = 5, d = 6
The nth term of AP is given by
Tn = a + (n-1) d
Tn= 5 + (n-1) x 6 = 5+ 6n – 6 = 6n – 1
(iv) The given AP is (5a – x), 6a, (7a + x) …..
Its first term = (5a – x) and common difference = 6a – 5a – x = a + x
The nth term of AP is given by
Tn = a + (n-1) d
T11 = (5a – x) + (11-1) (a + x)
= 5a – x + 10x + 10x
= 15a + 9x = 3(5a +3x)
Hence the 11th term is 3(5a + 3x)
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Question 4:
(i) The given AP is 63, 58, 53, 48 ….
First term = 63, common difference = 58 – 63 = – 5
∴ a = 63, d = – 5
The nth term of AP is given by
Tn = a + (n-1) d
T10 = 63 + (10-1) (-5) = 63- 45 = 18
Hence the 10th term is 18
(ii) The given AP is 9, 5, 1, -3….
First term = 9, common difference = 5 – 9 = -4
∴ a = 9, d= – 4
The nth term of AP is given by
Tn = a + (n-1) d
T14 = 9 + (14-1) (-4) = 9- 52 = -43
Hence, the 14th term is – 43
(iii) The given AP is 16, 9, 2, -5
First term = 16, common difference = 9 – 16 = – 7
∴ a = 16, d = -7
The nth term of AP is given by
Tn = a + (n-1) d
Tn = 16 + (n-1) (-7) ⇒ 16- 7n + 7 = (23 – 7n)
Hence, the nth term is (23 – 7n).
Question 5:
The given AP is \(6,7\frac { 3 }{ 4 } ,9\frac { 1 }{ 2 } ,11\frac { 1 }{ 4 } ……..\)
First term = 6, common difference = \(\left( 7\frac { 3 }{ 4 } -6 \right) \)
= \(\left( \frac { 31 }{ 4 } -6 \right) \)
= \(\frac { 7 }{ 4 } \)
a = 6, d = \(\frac { 7 }{ 4 } \)
The nth term is given by
Tn = a + (n-1) d
T14 = 6 + (37 – 1) \((\frac { 7 }{ 4 }) \) = 6+ 63 = 69
Hence, 37th term is 69
Question 6:
The given AP is \(5,4\frac { 1 }{ 2 } ,4,3\frac { 1 }{ 2 } ,3……..\)
The first term = 5,
common difference = \(\left( 4\frac { 1 }{ 2 } -5 \right) =\left( \frac { 9 }{ 2 } -5 \right) =-\frac { 1 }{ 2 } \)
∴ a = 5, d = \(-\frac { 1 }{ 2 } \)
The nth term is given by
Tn = a + (n-1) d
T14 = 5 + (25 – 1) (-1/2) = 5- 12 = -7
Hence the 25th term is – 7
Question 7:
In the given AP, we have a = 6 and d = (10 – 6) = 4
Suppose there are n terms in the given AP, then
Tn = 174 ⇒ a + (n-1) d = 174
⇒ 6 + (n-1) 4 = 174
⇒ 6 + 4n – 4 = 174
⇒ 2 + 4n = 174 ⇒ n = 172/4 ⇒ 43
Hence there are 43 terms in the given AP
Question 8:
In the given AP we have a = 41 and d = 38 – 41 = – 3
Suppose there are n terms in AP, then
Tn = 8 ⇒ a + (n-1) d = 8
⇒ 41 + (n-1) (-3) = 8
⇒ 41 – 3n + 3 = 8
⇒ -3n = – 36 ⇒ n = 12
Hence there are 12 terms in the given AP
Question 9:
In the given AP, we have a = 3 and d = 8 – 3 = 5
Suppose there are n terms in given AP, then
Tn = a + (n-1) d = 88
⇒ 3 + (n-1) 5 = 88
⇒ 3 + 5n – 5 = 88
⇒ 5n = 90
⇒ n = 12
Hence, the 18th term of given AP is 88
Question 10:
In the given AP, we have a = 72 and d = 68 – 72 = – 4
Suppose there are n terms in given AP, we have
Tn = 0 ⇒ a + (n-1) d = 0
⇒ 72 + (n-1) (-4) = 0
⇒ 72 – 4n + 4 = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19th term in the given AP is 0
Question 11:
In the given AP, we have a = \(\frac { 1 }{ 2 } \) ; \(\left( 1-\frac { 5 }{ 6 } \right) =\frac { 1 }{ 6 } \)
Suppose there are n terms in given AP, we have
Then,
Tn = 3 ⇒ a + (n-1) d = 3
⇒ \(\frac { 5 }{ 6 } +(n-1)\frac { 1 }{ 6 } =3\)
⇒ \(\frac { 5 }{ 6 } +\frac { 1 }{ 6 } n-\frac { 1 }{ 6 } =3\)
⇒ 4 + n = 18
⇒ n = 14
Thus, 14th term in the given AP is 3
Question 12:
We know that T1 – (5x + 2), T2 – (4x – 1) and T3 – (x + 2)
Clearly,
T2 – T1 = T3 – T2
⇒ (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 4x – 1 – 5x – 2 = x + 2 – 4x + 1
⇒ -x – 3 = -3x + 3
⇒ -x + 3x = 6
⇒ 2x = 6 ⇒ x = 3
Hence x = 3
Question 13:
Tn = (4n – 10)
⇒ T1 = (4 x 1 – 10) = -6 and T2 = (4 x 2 – 10) = -2
Thus, we have
(i) First term = -6
(ii) Common difference = (T2 – T1) = (-2+6) = 4
(iii) 16th term = a + (16-1) d, where a = -6 and d = 4
= (-6 + 15 x 4) = 54
Question 14:
In the given AP, let first term = a and common difference = d,
Then, Tn = a + (n-1) d
⇒ T4 = a + (4 – 1)d, T10 = a + (10 – 1)d
⇒ T4 = a + 3d, T10 = a + 9d
Now, T4 = 13 ⇒ a + 3d = 13 – – – (1)
T10 = 25 ⇒ a + 9d = 25 – – – (2)
Subtracting (1) from (2), we get
⇒ 6d = 12 ⇒ d = 2
Putting d = 2 in (1), we get
a + 3 x 2 = 13
⇒ a = (13 – 6) = 7
Tthus, a = 7, and d = 2
17th term = a + (17 – 1)d, where a= 7, d = 2
(7 + 16 x 2) = (7 + 32) = 39
∴ a = 7, d = 2,
Question 15:
In the given AP, let first term = a and common difference = d
Then, Tn = a + (n-1) d
⇒ T8 = a + (8 – 1)d, T12 = a + (12 – 1)d
⇒ T8 = a + 7d, T12 = a + 11d
Now, T8 = 37 ⇒ a + 7d = 37 – – – (1)
T12 = 57 ⇒ a + 11d = 57 – – – (2)
Subtracting (1) from (2), we get
⇒ 4d = 20 ⇒ d = 5
Putting d = 5 in (1), we get
a + 7 x 5 = 37
⇒ a = 2
Tthus, a = 2, and d = 5
So the required AP is 2, 7, 12..
Question 16:
In the given AP, let the first term = a, and common difference = d
Then, Tn = a + (n-1) d
⇒ T7 = a + (7 – 1)d, and T13 = a + (13 – 1)d
⇒ T7 = a + 6d, T13 = a + 12d
Now, T7 = -4 ⇒ a + 6d = -4 – – – (1)
T13 = -16 ⇒ a + 12d = -16 – – – (2)
Subtracting (1) from (2), we get
⇒ 6d = -12 ⇒ d = -2
Putting d = -2 in (1), we get
a + 6 (-2) = -4
⇒ a – 12 = -4
⇒ a = 8
Tthus, a = 8, and d = -2
So the required AP is 8, 6, 4, 2, 0……
Question 17:
In the given AP let the first term = a, And common difference = d
Then, Tn = a + (n-1) d
⇒ T10 = a + (10 – 1)d, T17 = a + (17 – 1)d, T13 = a + (13 – 1)d
⇒ T10 = a + 9d, T17 = a + 16d, T13 = a + 12d
Now, T10 = 52 ⇒ a + 9d = 52 – – – (1)
and T17 = T13 + 20 ⇒ a + 16d = a + 12d + 20
⇒ 4d = 20 ⇒ d = 5
Putting d = 5 in (1), we get
a + 9 x 5 = 52 ⇒ a = 52-45 ⇒ a = 7
Thus, a = 7 and d = 5
So the required AP is 7, 12, 17, 22….
Question 18:
Let the first term of given AP = a and common difference = d
Then, Tn = a + (n-1) d
⇒ T4 = a + (4 – 1)d, T25 = a + (25 – 1)d, T11 = a + (11 – 1)d
⇒ T4 = a + 3d, T25 = a + 24d, T11 = a + 10d
Now, T4 = 0 ⇒ a + 3d = 0 ⇒ a = -3d
∴ T25 = a + 24d = (-3d +24d) ⇒ 21d
and T11 = a + 10d = (-3d +10d) ⇒ 7d
∴ T25 = 21d = 3 x 7d = 3 x T11
Hence 25th term is triple its 11th term
Question 19:
The given AP is 3, 8, 13, 18…..
First term a = 3, common difference a = 8 – 3 = 5
∴ Tn = a + (n-1) d = 3 + (n – 1) x 5 = 5n – 2
T20 = 3 + (20-1) 5 = 3 + 19 x 5 = 98
Let nth term is 55 more than the 20th term
∴ (5n – 2) – 98 = 55
Or 5n = 100 + 55 = 155
n = 155/5 = 31
∴ 31st term is 55 more than the 20th term of given AP
Question 20:
The given AP is 5, 15, 25….
a = 5, d = 15 – 5 = 10
We have, Tn = 130+T31
⇒ a + (n-1) d = 130 + 5 + (31 – 1) x 10
⇒ 5 + (n-1) 10 = 130 + 5 + (31 – 1) x 10
⇒ 5 + 10n – 10 = 135 + 300
⇒ 10n – 5 = 435 or 10n = 453 + 5
∴ n = 440/10 = 44
Thus, the required term is 44th
Question 21:
First AP is 63, 65, 67….
First term = 63, common difference = 65 – 63 = 2
∴ nth term = 63 + (n – 1) 2 = 63 + 2n – 2 = 2n + 61
Second AP is 3, 10, 17 ….
First term = 3, common difference = 10 – 3 = 7
nth term = 3 + (n – 1) 7 = 3 + 7n – 7 = 7n – 4
The two nth terms are equal
∴ 2n + 61 = 7n – 4 or 5n = 61 + 4 = 65
⇒ n = 65/4 = 13.
Question 22:
Three digit numbers which are divisible by 7 are 105, 112, 119,….994
This is an AP where a= 105, d = 7 and l = 994
Let nth term be 994
∴ a + (n – 1)d =994 or 105 + (n – 1)7 = 994
⇒ 105 + 7n – 7 = 994 or 7n = 94 – 98 = 896
∴ n = 896/7 = 128.
Hence, there are 128 three digits number which are divisible by 7.
Question 23:
Here a = 7, d = (10 – 7) = 3, l = 184
And n = 8
Now, nth term from the end = [ l – (n-1) d ]
= [ 184 – (8-1) 3 ]
= [ 184 – 7 x 3]
= 184-21
= 163
Hence, the 8th term from the end is 163
Question 24:
Here a = 17, d = (14 – 17) = -3, l = -40
And n = 6
Now, nth term from the end = [ l – (n – 1) d ]
= [ -40 – (6-1)(-3) ]
= [ -40 + 5 x 3]
= -40+15
= -25
Hence, the 6th term from the end is – 25
Question 25:
The given AP is 10, 7, 4, ….. (-62)
a = 10, d = 7 – 10 = -3, l = -62
Now, 11th term from the end = [ l – (n – 1) d ]
= [ -62 – (11-1)(-3) ]
= -62 + 30
= -32
Question 26:
Let a be the first term and d be the common difference
pth term = a +(p – 1)d = q (given) —–(1)
qth term = a +(q – 1) d = p (given) —–(2)
subtracting (2) from (1)
(p – q)d = q – p
(p – q)d = -(p – q)
d = -1
Putting d = -1 in (1)
a – (p – 1) = q ∴ a = p + q -1
∴ (p + q)th term = a+ (p + q -1)d
= (p + q -1) – (p + q -1) = 0
Question 27:
Let a be the first term and d be the common difference
T10 = a + 9d, T15 = a + 14d
10T10 = 15T15
⇒ 10(a + 9) d = 15(a + 14d)
⇒ 2(a + 9) d = 3(a + 14d)
⇒ a + 24d = 0
∴ T25 = 0
Question 28:
Let a be the first term and d be the common difference
∴ nth term from the beginning = a + (n – 1)d —–(1)
nth term from end = l – (n – 1)d —-(2)
adding (1) and (2),
sum of the nth term from the beginning and nth term from the end = [a + (n – 1)d] + [l – (n – 1)d] = a + l
Question 29:
Number of rose plants in first, second, third rows…. are 43, 41, 39… respectively.
There are 11 rose plants in the last row
So, it is an AP . viz. 43, 41, 39 …. 11
a = 43, d = 41 – 43 = -2, l = 11
Let nth term be the last term
∴ l = a + (n-1) d
⇒ 11 = 43 + (n-1) x (-2)
43 – 2n + 2 = 11 or 2n = 45 -11 = 34
∴ n = 34/2 = 17
Hence, there are 17 rows in the flower bed.
Question 30:
Total amount = ₹ 2800
and number of prizes = 4
Let first prize = ₹ a
Then second prize = ₹ a – 200
Third prize = a – 200 – 200 = a – 400
and fourth prize = a – 400 – 200 = a – 600
But sum of there 4 prizes are ₹ 2800
a + a – 200 + a – 400 + a – 600 = ₹ 2800
⇒ 4a – 1200 = 2800
⇒ 4a = 2800 + 1200 = 4000
⇒ a = 1000
First prize = ₹ 1000
Second prize = ₹ 1000 – 200 = ₹ 800
Third prize = ₹ 800 – 200 = ₹ 600
and fourth prize = ₹ 600 – 200 = ₹ 400
Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.
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