## RS Aggarwal Solutions Class 10 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

**Exercise 11D**

**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

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

S_{1} = 2(1)² + 3 x 1 = 2 + 3 = 5

S_{2} = 2(2)² + 3 x 2 = 8 + 6=14

S_{3} = 2(3)² + 3 x 3 = 18 + 9 = 27

Now, T_{2} = S_{2} – S_{1} = 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, a_{30} = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147

and a_{20} = 2 + (20 – 1) x 5 = 2 + 19 x 5 = 2 + 95 = 97

a_{30} – a_{20} = 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

T_{10} = 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 = T_{2} – T_{1} = 14 – 8 = 6

Common difference = 6

**Question 25.**

**Solution:**

**Question 26.**

**Solution:**

**Question 27.**

**Solution:**

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

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