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

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D

Other Exercises

Very-Short and Short-Answer Questions
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
Tn = (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(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
Sm = 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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 1

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
an = 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:
Tn = 3n + 5
Tn-1 = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2
d = Tn – Tn-1 = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3
Common difference = 3

Question 13.
Solution:
T= 7 – 4n
Tn-1 = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n
d = Tn – Tn-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), ………
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 2

Question 15.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 3

Question 16.
Solution:
AP is 21, 18, 15, …n
Here, a = 21, d = 18 – 21 = -3, l = 0
Tn (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
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 4
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 5

Question 18.
Solution:
First n even natural numbers are 2, 4, 6, 8, 10, … n
Here, a = 2, d = 4 – 2 = 2
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 6

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:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 7

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:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 8

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 9
d = T2 – T1 = 14 – 8 = 6
Common difference = 6

Question 25.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 10

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 11

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 12
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11D 13

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