RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11D Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 11 Arithmetic Progressions Ex 11D Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D Maths book pdf download. Now you will get step by step solution to each question.

RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D PDF

Question 1.
Solution:
(3y – 1), (3y + 5) and (5y+ 1) are in AP
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
⇒ 2 (3y + 5) = (5y + 1) + (3y – 1)
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 2.
Solution:
k, (2k – 1) and (2k + 1) are the three successive terms of an AP.
(2k – 1) – k = (2k + 1) – (2k – 1)
⇒ 2 (2k – 1) = 2k + 1 + k
⇒ 4k – 2 = 3k + 1
⇒ 4k – 3k = 1 + 2
⇒ k = 3
k = 3

Question 3.
Solution:
18, a, (b – 3) are in AP
⇒ a – 18 = b – 3 – a
⇒ a + a – b = -3 + 18
⇒ 2a – b = 15

Question 4.
Solution:
a, 9, b, 25 are in AP.
9 – a = b – 9 = 25 – b
b – 9 = 25 – b
⇒ b + b = 22 + 9 = 34
⇒ 2b = 34
⇒ b= 17
a – b = a – 9
⇒ 9 + 9 = a + b
⇒ a + b = 18
⇒ a + 17 = 18
⇒ a = 18 – 17 = 1
a = 18, b= 17

Question 5.
Solution:
(2n – 1), (3n + 2) and (6n – 1) are in AP
(3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)
⇒ (3n + 2) + (3n + 2) = 6n – 1 + 2n – 1
6n + 4 = 8n – 2
⇒ 8n – 6n = 4 + 2
⇒ 2n = 6
⇒ n = 3
and numbers are
2 x 3 – 1 = 5
3 x 3 + 2 = 11
6 x 3 – 1 = 17
i.e. (5, 11, 17) are required numbers.

Question 6.
Solution:
Three digit numbers are 100 to 990 and numbers which are divisible by 7 will be
105, 112, 119, 126, …, 994
Here, a = 105, d= 7, l = 994
T_n = (l) = a + (n – 1) d
⇒ 994 = 105 + (n – 1) x 7
⇒ 994 – 105 = (n – 1) 7
⇒ (n – 1) x 7 = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
Required numbers are 128

Question 7.
Solution:
Three digit numbers are 100 to 999
and numbers which are divisible by 9 will be
108, 117, 126, 135, …, 999
Here, a = 108, d= 9, l = 999
T_n (l) = a + (n – 1) d
⇒ 999 = 108 + (n – 1) x 9
⇒ (n – 1) x 9 = 999 – 108 = 891
⇒ n – 1 = 99
⇒ n = 99 + 1 = 100

Question 8.
Solution:
Sum of first m terms of an AP = 2m² + 3m
S_m = 2m² + 3m
S₁ = 2(1)² + 3 x 1 = 2 + 3 = 5
S₂ = 2(2)² + 3 x 2 = 8 + 6=14
S₃ = 2(3)² + 3 x 3 = 18 + 9 = 27
Now, T₂ = S₂ – S₁ = 14 – 5 = 9
Second term = 9

Question 9.
Solution:
AP is a, 3a, 5a, …
Here, a = a, d = 2a

Question 10.
Solution:
AP 2, 7, 12, 17, …… 47
Here, a = 2, d = 7 – 2 = 5, l = 47
nth term from the end = l – (n – 1 )d
5th term from the end = 47 – (5 – 1) x 5 = 47 – 4 x 5 = 47 – 20 = 27

Question 11.
Solution:
AP is 2, 7, 12, 17, …
Here, a = 2, d = 7 – 2 = 5
a_n = a + (n – 1) d = 2 + (n – 1) x 5 = 2 + 5n – 5 = 5n – 3
Now, a₃₀ = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147
and a₂₀ = 2 + (20 – 1) x 5 = 2 + 19 x 5 = 2 + 95 = 97
a₃₀ – a₂₀ = 147 – 97 = 50

Question 12.
Solution:
T_n = 3n + 5
T_n-1 = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2
d = T_n – T_n-1 = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3
Common difference = 3

Question 13.
Solution:
T_n = 7 – 4n
T_n-1 = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n
d = T_n – T_n-1 = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4
d = -4

Question 14.
Solution:
AP is √8, √18, √32, …..
⇒ √(4 x 2) , √(9 x 2) , √(16 x 2), ………

Question 15.
Solution:

Question 16.
Solution:
AP is 21, 18, 15, …n
Here, a = 21, d = 18 – 21 = -3, l = 0
T_n (l) = a + (n – 1) d
0 = 21 + (n – 1) x (-3)
0 = 21 – 3n + 3
⇒ 24 – 3n = 0
⇒ 3n = 24
⇒ n = 8 .
0 is the 8th term.

Question 17.
Solution:
First n natural numbers are 1, 2, 3, 4, 5, …, n
Here, a = 1, d = 1

Question 18.
Solution:
First n even natural numbers are 2, 4, 6, 8, 10, … n
Here, a = 2, d = 4 – 2 = 2

Question 19.
Solution:
In an AP
First term (a) = p
and common difference (d) = q
T₁₀ = a + (n – 1) d = p + (10 – 1) x q = (p + 9q)

Question 20.
Solution:

Question 21.
Solution:
2p + 1, 13, 5p – 3 are in AP, then
13 – (2p + 1) = (5p – 3) – 13
⇒ 13 – 2p – 1 = 5p – 3 – 13
⇒ 12 – 2p = 5p – 16
⇒ 5p + 2p = 12 + 16
⇒ 7p = 28
⇒ p = 4
P = 4

Question 22.
Solution:
(2p – 1), 7, 3p are in AP, then
⇒ 7 – (2p – 1) = 3p – 7
⇒ 7 – 2p + 1 = 3p – 7
⇒ 7 + 1 + 7 = 3p + 2p
⇒ 5p = 15
⇒ p = 3
P = 3

Question 23.
Solution:

Question 24.
Solution:

d = T₂ – T₁ = 14 – 8 = 6
Common difference = 6

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

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