# RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A

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### RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11APDF

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

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
T= 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
T= 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
T= 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
T= 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)

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
T= 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
T= 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
T= 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   6,734,912,1114……..
First term = 6, common difference =   (734−6)
=   (314−6)
=   74
a = 6, d =   74
The nth term is given by
T= a + (n-1) d
T14 = 6 + (37 – 1)  (74) = 6+ 63 = 69
Hence, 37th term is 69

Question 6:
The given AP is     5,412,4,312,3……..
The first term = 5,
common difference = (412−5)=(92−5)=−12
∴ a = 5, d = −12
The nth term is given by
T= 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
T = 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
T = 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
T =  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
T = 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 = 12 ; (1−56)=16
Suppose there are n terms in given AP, we have
Then,
T = 3 ⇒ a + (n-1) d = 3
⇒  56+(n−1)16=3
⇒  56+16n−16=3
⇒ 4 + n = 18
⇒ n = 14
Thus, 14th term in the given AP is 3
Question 12:
We know that   T– (5x + 2), T– (4x – 1) and  T– (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:
T = (4n – 10)
⇒ T = (4 x 1 – 10) = -6  and  T = (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, T =  a + (n-1) d
⇒ T = a + (4 – 1)d, T10  = a + (10 – 1)d
⇒ T = a + 3d, T10  = a + 9d
Now, T = 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, T =  a + (n-1) d
⇒ T = a + (8 – 1)d, T12  = a + (12 – 1)d
⇒ T = a + 7d, T12  = a + 11d
Now, T = 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, T =  a + (n-1) d
⇒ T = a + (7 – 1)d, and T13  = a + (13 – 1)d
⇒ T = a + 6d, T13  = a + 12d
Now, T = -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, T =  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, T =  a + (n-1) d
⇒ T = a + (4 – 1)d, T25  = a + (25 – 1)d, T11  = a + (11 – 1)d
⇒ T = a + 3d, T25  = a + 24d, T11  = a + 10d
Now, T = 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
∴  T =  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,  T =  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)
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

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