# RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11C Class 10 Maths

 Chapter Name RS Aggarwal Chapter 11 Arithmetic Progression Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 11AExercise 11BExercise 11D Related Study NCERT Solutions for Class 10 Maths

### Exercise 11C Solutions

1. The first three terms of an AP are respectively (33y – 1), (3y + 5) and (5y + 1), find the value of y.

Solution

The terms (3y – 1), (3y + 5) and (5y + 1) are in AP.

∴ (3y + 5) - (3y – 1) = (5y + 1) – (3y + 5)

⇒ 3y + 5 – 3y + 1= 5y + 1 – 3y – 5

⇒ 6 = 2y – 4

⇒ 2y = 10

⇒ y = 5

Hence, the value of y is 5.

2. If k, (2k – 1) and (2k + 1) are the three successive terms of an AP, find the value of k.

Solution

It is given that k, (2k – 1) and (2k + 1) are the three successive terms of an AP.

∴ (2k – 1) – k = (2k + 1) – (2k – 1)

⇒ k – 1 = 2

⇒ k = 3

Hence, the value of k is 3.

3. If 18, a, (b – 3) are in AP, then find the value of (2a – b)

Solution

It is given that 18, a, (b – 3) are in AP.

∴ a – 18 = (b – 3) – a

⇒ a + a – b = 18 – 3

⇒ 2a – b = 15

Hence, the required value is 15.

4. If the numbers a, 9, b, 25 from an AP, find a and b.

Solution

It is given that the numbers a, 9, b, 25 from an AP.

∴ 9 – a = b – 9 = 25 – b

So,

b – 9 = 25 – b

⇒ 2b = 34

⇒ b = 17

Also,

9 – a = b – 9

⇒ a = 18 – b

⇒ a = 18 – 17 (b = 17)

⇒ a = 1

Hence, the required value of a and b are 1 and 17, respectively.

5. If the numbers (2n – 1), (3n + 2) and (6n – 1) are in AP, find the value of n and the numbers.

Solution

It is given that numbers (2n – 1), (3n + 2) and (6n – 1) are in AP.

∴ (3n + 2) – (2n – 1) = (6n – 1 – (3n + 2)

⇒ 3n + 2 – 2n + 1 = 6n – 1 – 3n – 2

⇒ n + 3 = 3n – 3

⇒ 2n = 6

⇒ n = 3

When, n = 3

2n – 1 = 2×3 – 1 = 6 – 1 = 5

3n + 2 = 3×3 + 2 = 9 + 2 = 11

6n – 1 = 6×3 – 1 = 18 – 1 = 17

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.

6. How many three-digit natural numbers are divisible by 7?

Solution

The three-digit natural numbers divisible by 7 are 105, 112, 119, .......,994

Clearly, these number are in AP.

Here, a = 105 and d = 112 – 105 = 7

Let this AP contains n terms. Then,

an = 994

⇒ 105 + (n – 1) × 7 = 994 [an = a + (n – 1)d]

⇒ 7n + 98 = 994

⇒ 7n = 994 – 98 = 986

⇒ n = 128

Hence, there are 128 three-digit numbers divisible by 7.

7. How many three-digit natural numbers are divisible by 9?

Solution

The three-digit natural numbers divisible by 9 are 108, 117, 126, ......999.

Clearly, these number are in AP.

Here, a = 108 and d = 117 – 108 = 9

Let this AP contains n terms. Then,

an = 999

⇒ 108 + (n – 1) × 9 = 999 [an = a + (n – 1)d]

⇒ 9n + 99 = 999

⇒ 9n = 999 – 99 = 900

⇒ n = 100

Hence, there are 100 three-digit numbers divisible by 9.

8. If the sum of first m terms of an AP is (2m2 + 3m) then what is its second term?

Solution

Let Sm denotes the sum of first m terms of the AP.

∴ Sm = 2m2 + 3m

⇒ Sm-1 = 2(m – 1)2 + 3(m – 1) = 2(m2 – 2m + 1) + 3(m – 1) = 2m2 – m – 1

Now,

mth term of AP, am = Sm – Sm-1

∴ am = (2m2 + 3m) – (2m2 – m – 1) = 4m + 1

Putting m = 2, we get

a2 = 4 × 2 + 1 = 9

Hence, the second term of the AP is 9.

9. What is the sum of first n terms of the AP a, 3a, 5a, .......

Solution

The given AP is 3a, 5a, .......

Here,

First term, A = a

Common difference, D = 3a – a = 2a

∴ Sum of the n terms, Sn

= n/2 [2 × a + (n – 1) × 2a] {Sn = n/2[2A + (n – 1)D]}

= n/2 (2a + 2an – 2a)

= n/2 × 2an

= an2

Hence, the required sum is an2.

10. What is the 5th term from the end of the AP 2, 7, 12, ......,47?

Solution

The given AP is 2, 7, 12, ....., 47.

Let us re-write the given AP in reverse order i.e., 47, 42, ....12, 7, 2.

Now, the 5th term from the end of the given AP is equal to the 5th term from the beginning of the AP 47, 42, ...., 12, 7, 2.

Consider the AP 47, 42, ....12, 7, 2.

Here, a = 47 and d = 42 – 47 = - 5

5th term of this AP

= 47 + (5 – 1) × (-5)

= 47 – 20

= 27

Hence, the 5th term from the end of the given AP is 27.

11. If an denotes the nth term of the AP 2, 7, 12, 17, .....find the value of (a30 – a20).

Solution

The given AP is 2, 7, 12, 17, ......

Here, a = 2 and d = 7 – 2 = 5

∴ a30 – a20

= [2 + (30 – 1) × 5] – [2 + (20 – 1) × 5] [an = a + (n – 1)d]

= 147 – 97

= 50

Hence, the required value is 50.

12. The nth term of an AP is (3n + 5). Find its common difference.

Solution

We have

Tn = (3n + 5)

Common difference = T2 – T1

T1 = 3 × 1 + 5 = 8

T2 = 3 × 2 + 5 = 11

d = 11 – 8 – 3

Hence, the common difference is 3.

13. The nth term of an AP is (7 – 4n). Find its common difference.

Solution

We have

Tn = (7 – 4n)

Common difference = T2 – T1

T1 = 7 - 4 × 1 = 3

T2 = 7 – 4 × 2 = - 1

d = - 1 – 3 = -4

Hence, the common difference is -4.

14. Write the next term for the AP ✓8, ✓18, ✓32 ....

Solution

15. Write the next term for the AP ✓2, ✓8, ✓18 ...

Solution

16. Which term of the AP 21, 18, 15, .... is zero?

Solution

In the given AP, first term a = 21 and common difference, d = (18 – 21) = - 3

Let’s its nth term be 0.

Then, Tn = 0

⇒ a + (n – 1)d = 0

⇒ 21 + (n – 1) × (-3) = 0

⇒ 24 – 3n = 0

⇒ 3n = 24

⇒ n = 8

Hence, the 8th term of the given AP is 0.

17. Find the sum of the first n natural numbers.

Solution

The first n natural numbers 1, 2, 3, 4, 5, ....,n

Here, a = 1 and d = (2 – 1) = 1

Sum of n terms of an AP is given by

Sn = n/2[2a + (n – 1)d]

= (n/2) × [2 × 1 + (n – 1) × 1]

= (n/2) × [2 + n – 1]

= (n/2) × (n + 1)

= n(n + 1)/2

18. Find the sum of first n even numbers.

Solution

The first n even natural numbers are 2, 4, 6, 8, 10, ......, n.

Here, a = 2 and d = (4 – 2) = 2

Sum of n terms of an AP is given by

Sn = n/2[2a + (n – 1)d]

= (n/2) × [2 × 2 + (n – 1) × 2]

= (n/2) × [4 + 2n – 2]

= (n/2) × (2n + 2)

= n(n + 1)

Hence, the required sum is n(n + 1).

19. The first term of an AP is p and its common difference is q. Find its 10th term.

Solution

Here, a = p and d = q

Now, Tn = a + (n – 1)d

⇒ Tn = p + (n – 1)q

∴ T10 = p + 9q

20. If (2p + 1), 13, (5p – 3) are in AP, find the value of p.

Solution

Let (2p + 1), 13, (5p – 3) be three consecutive terms of an AP.

Then 13 – (2p + 1) = (5p – 3) – 13

⇒ 7p = 28

⇒ p = 4

∴ When p = 4, (2p + 1), 13 and (5p – 3) from three consecutive terms of an AP.

21. If (2p – 1), 7, 3p are in AP, find the value of p.

Solution

Let (2p – 1), 7 and 3p be three consecutive terms of an AP.

Then 7 – (2p – 1) = 3p – 7

⇒ 5p = 15

⇒ p = 3

∴ When p = 3, (2p – 1), 7 and 3p from three consecutive terms of an AP.

22. If the sum of first p terms of an AP is (ap2 + bp), find its common difference.

Solution

Let Sp denotes the sum of first p terms of the AP.

∴ Sp = ap2 + bp

⇒ Sp –1 = a(p – 1)2 + b(p – 1)

= a(p2 – 2p + 1) + b(p – 1)

= ap2 – (2a – b)p + (a – b)

Now,

pth term of AP, ap = Sp – Sp-1

= (ap2 + bp) – [ap2 – (2a – b)p + (a – b)]

= ap2 + bp – ap2 + (2a – b)p + (a – b)]

= 2ap – (a – b)

Let d be the common difference of the AP.

∴ d = ap – ap-1

= [2ap – (a – b)]

= [2a(p – 1) – (a – b)]

= 2ap – (a – b) – 2a(p – 1) + (a – b)

= 2a

Hence, the common difference of the AP is 2a.

23. If the sum of first n terms is (3n2 + 5n), find its common difference.

Solution

Let Sn denotes the sum of first n terms of the AP.

∴ Sn = 3n2 + 5n

⇒ Sn-1 = 3(n – 1)2 + 5(n – 1)

= 3(n2 – 2n + 1) + 5(n – 1)

= 3n2 – n – 2

Now,

nth term of AP, an = Sn – Sn-1

= (3n2 + 5n) – (3n2 – n – 2)

= 6n + 2

Let d be the common difference of the AP.

∴ d = an – an-1

= (6n + 2) – [6(n – 1) + 2]

= 6n + 2 – 6(n – 1) – 2

= 6

Hence, the common difference of the AP is 6.

24. Find the AP whose 4th term is 9 and thee sum of its 6th and 13th terms is 40.

Solution

Let a be the first term and d be the common difference of the AP. Then,

a4 = 9

⇒ a + (4 – 1)d = 9 [an = a + (n – 1)d]

⇒ a + 3d = 9 ...(1)

Now,

a6 + a13 = 40 (Given)

⇒ (a + 5d) + (a + 12d) = 40

⇒ 2a + 17d = 40 ...(2)

From (1) and (2), we get

2(9 – 3d) + 17d = 40

⇒ 18 – 6d + 17d = 40

⇒ 11d = 40 – 18 = 22

⇒ d = 2

Putting d = 2 in (1), we get

a + 3 × 2 = 9

⇒ a = 9 – 6 = 3

Hence, the AP is 3, 5, 7, 9, 11, ....