RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression | Updated For 2024

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RS Aggarwal Solutions Class 10 Maths Chapter 11 Arithmetic Progression

 

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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 n^th term of the AP is given by
T_n = a + (n-1) d
T₂₀ = 1 + (20-1) x 4 = 1+ 76 = 77
Hence, the 20^th 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 n^th term of the AP is given by
T_n = a + (n-1) d
T35 = 6 + (35-1) x 3 = 6+ 102 = 108
Hence, the 35^th 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 n^th term of AP is given by
T_n = a + (n-1) d
T_n= 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 n^th term of AP is given by
T_n = a + (n-1) d
T₁₁ = (5a – x) + (11-1) (a + x)
= 5a – x + 10x + 10x
= 15a + 9x = 3(5a +3x)
Hence the 11^th 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 n^th term of AP is given by
T_n = a + (n-1) d
T₁₀ = 63 + (10-1) (-5) = 63- 45 = 18
Hence the 10^th 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 n^th term of AP is given by
T_n = a + (n-1) d
T₁₄ = 9 + (14-1) (-4) = 9- 52 = -43
Hence, the 14^th 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 n^th term of AP is given by
T_n = a + (n-1) d
T_n = 16 + (n-1) (-7) ⇒ 16- 7n + 7 = (23 – 7n)
Hence, the n^th 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 n^th term is given by
T_n = a + (n-1) d
T₁₄ = 6 + (37 – 1)  (74) = 6+ 63 = 69
Hence, the 37^th 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 n^th term is given by
T_n = a + (n-1) d
T₁₄ = 5 + (25 – 1) (-1/2) = 5- 12 = -7
Hence the 25^th 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_n  = 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_n  = 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 a given AP, then
T_n  =  a + (n-1) d = 88
⇒ 3 + (n-1) 5 = 88
⇒ 3 + 5n – 5 = 88
⇒ 5n = 90
⇒  n = 12
Hence, the 18^th term of the 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 a given AP, we have
T_n  = 0 ⇒ a + (n-1) d = 0
⇒ 72 + (n-1) (-4) = 0
⇒ 72 – 4n + 4 = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19^th 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 a given AP, we have
Then,
T_n  = 3 ⇒ a + (n-1) d = 3
⇒  56+(n−1)16=3
⇒  56+16n−16=3
⇒ 4 + n = 18
⇒ n = 14
Thus, 14^th 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:
T_n  = (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) 16^th term = a + (16-1) d, where a = -6 and d = 4
= (-6 + 15 x 4) = 54
Question 14:
In the given AP, let the first term = a and common difference = d,
Then, T_n  =  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
Thus, a = 7, and d = 2
17^th 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 the first term = a and common difference = d
Then, T_n  =  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
Thus, 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_n  =  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
Thus, 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_n  =  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_n  =  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 25^th term is triple its 11^th term
Question 19:
The given AP is 3, 8, 13, 18…..
First term a = 3, common difference a = 8 – 3 = 5
∴  T_n  =  a + (n-1) d = 3 + (n – 1) x 5 = 5n – 2
T₂₀  =  3 + (20-1) 5 = 3 + 19 x 5 = 98
Let the n^th  term is 55 more than the 20^th term
∴  (5n – 2) – 98 = 55
Or 5n = 100 + 55 = 155
n = 155/5 = 31
∴  31^st term is 55 more than the 20^th term of the given AP
Question 20:
The given AP is 5, 15, 25….
a = 5, d = 15 – 5 = 10
We have,  T_n  =  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 44^th
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
The 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 that are divisible by 7 are 105, 112, 119,….994
This is an AP where a= 105, d = 7 and l = 994
Let the n^th 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 numbers that 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 8^th term from the end is 163
Question 24:
Here a = 17, d = (14 – 17) = -3, l = -40
And n = 6
Now, an n^th term from the end = [ l – (n – 1) d ]
= [ -40 – (6-1)(-3) ]
= [ -40 + 5 x 3]
= -40+15
= -25
Hence, the 6^th 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, the 11^th 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
p^th term = a +(p – 1)d = q (given) —–(1)
q^th 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
∴  n^th term from the beginning = a + (n – 1)d —–(1)
n^th term from end = l – (n – 1)d —-(2)
adding (1) and (2),
the sum of the n^th term from the beginning and the n^th term from the end = [a + (n – 1)d] + [l – (n – 1)d] = a + l
Question 29:
The number of rose plants in the first, second, and third rows…. is 43, 41, and 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 the n^th 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
The 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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