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 
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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 threedigit natural numbers are divisible by 7?
Solution
The threedigit 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,
a_{n} = 994
⇒ 105 + (n – 1) × 7 = 994 [a_{n} = a + (n – 1)d]
⇒ 7n + 98 = 994
⇒ 7n = 994 – 98 = 986
⇒ n = 128
Hence, there are 128 threedigit numbers divisible by 7.
7. How many threedigit natural numbers are divisible by 9?
Solution
The threedigit 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,
a_{n} = 999
⇒ 108 + (n – 1) × 9 = 999 [a_{n} = a + (n – 1)d]
⇒ 9n + 99 = 999
⇒ 9n = 999 – 99 = 900
⇒ n = 100
Hence, there are 100 threedigit numbers divisible by 9.
8. If the sum of first m terms of an AP is (2m^{2 }+ 3m) then what is its second term?
Solution
Let S_{m} denotes the sum of first m terms of the AP.
∴ S_{m} = 2m^{2} + 3m
⇒ S_{m1} = 2(m – 1)^{2} + 3(m – 1) = 2(m^{2} – 2m + 1) + 3(m – 1) = 2m^{2} – m – 1
Now,
m^{th} term of AP, am = S_{m} – S_{m1}
∴ a_{m} = (2m^{2} + 3m) – (2m^{2} – m – 1) = 4m + 1
Putting m = 2, we get
a_{2} = 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, S_{n}
= n/2 [2 × a + (n – 1) × 2a] {S_{n} = n/2[2A + (n – 1)D]}
= n/2 (2a + 2an – 2a)
= n/2 × 2an
= an^{2}
Hence, the required sum is an^{2}.
10. What is the 5^{th} term from the end of the AP 2, 7, 12, ......,47?
Solution
The given AP is 2, 7, 12, ....., 47.
Let us rewrite the given AP in reverse order i.e., 47, 42, ....12, 7, 2.
Now, the 5^{th} term from the end of the given AP is equal to the 5^{th} 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
5^{th} term of this AP
= 47 + (5 – 1) × (5)
= 47 – 20
= 27
Hence, the 5^{th} term from the end of the given AP is 27.
11. If a_{n} denotes the nth term of the AP 2, 7, 12, 17, .....find the value of (a_{30 }– a_{20}).
Solution
The given AP is 2, 7, 12, 17, ......
Here, a = 2 and d = 7 – 2 = 5
∴ a_{30 }– a_{20}
= [2 + (30 – 1) × 5] – [2 + (20 – 1) × 5] [a_{n} = 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
T_{n} = (3n + 5)
Common difference = T_{2} – T_{1 }
T_{1} = 3 × 1 + 5 = 8
T_{2 }= 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
T_{n} = (7 – 4n)
Common difference = T_{2} – T_{1 }
T_{1 }= 7  4 × 1 = 3
T_{2} = 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
Solution
In the given AP, first term a = 21 and common difference, d = (18 – 21) =  3
Let’s its n^{th} term be 0.
Then, T_{n} = 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
S_{n} = 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
S_{n} = 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 10^{th} term.
Solution
Here, a = p and d = q
Now, T_{n }= a + (n – 1)d
⇒ T_{n} = p + (n – 1)q
∴ T_{10} = 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 (ap^{2} + bp), find its common difference.
Solution
Let S_{p }denotes the sum of first p terms of the AP.
∴ S_{p} = ap^{2} + bp
⇒ S_{p –1 }= a(p – 1)^{2} + b(p – 1)
= a(p^{2} – 2p + 1) + b(p – 1)
= ap^{2} – (2a – b)p + (a – b)
Now,
p^{th} term of AP, ap = S_{p} – S_{p1}
= (ap^{2} + bp) – [ap^{2} – (2a – b)p + (a – b)]
= ap^{2} + bp – ap^{2} + (2a – b)p + (a – b)]
= 2ap – (a – b)
Let d be the common difference of the AP.
∴ d = a_{p} – a_{p1}
= [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 (3n^{2 }+ 5n), find its common difference.
Solution
Let S_{n} denotes the sum of first n terms of the AP.
∴ S_{n} = 3n_{2} + 5n
⇒ S_{n1} = 3(n – 1)^{2} + 5(n – 1)
= 3(n^{2} – 2n + 1) + 5(n – 1)
= 3n^{2} – n – 2
Now,
n^{th} term of AP, an = S_{n} – S_{n1}
= (3n^{2 }+ 5n) – (3n^{2} – n – 2)
= 6n + 2
Let d be the common difference of the AP.
∴ d = a_{n} – a_{n1}
= (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 4^{th} term is 9 and thee sum of its 6^{th} and 13^{th} terms is 40.
Solution
Let a be the first term and d be the common difference of the AP. Then,
a_{4} = 9
⇒ a + (4 – 1)d = 9 [a_{n} = a + (n – 1)d]
⇒ a + 3d = 9 ...(1)
Now,
a_{6} + a_{13} = 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, ....