# Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions

Q1. How many two-digit numbers are divisible by 7?
Sol. Let 'n' two- digit numbers are divisible by 7.
So, the numbers are: 14, 21, 28, 35, 42, 49 ....................98.
This series form an A.P.
∴ a = 14,d = 7,and   an = 98
Now, using   an = a + (n - 1) d, we get
98 = 14 + (n - 1) 7
⇒ 98 = 14 + 7n - 7
⇒ 98 = 7 + 7n  or  7n
= 98 - 7 = 91 ⇒ n = 91/7  = 13
Thus, required two-digit numbers is 13

Q2. If the numbers x - 2, 4x - 1 and 5x + 2 are in A.P. Find the value of x.
Sol.
∵ x - 2, 4x - 1 and 5x + 2 are in A.P.
∴ (4x - 1) - (x - 2) = (5x + 2) - (4x - 1)
⇒ 3x + 1 = x + 3
⇒ 2x = 2  ⇒ x = 1

Q3. Which term of the A.P. 4, 9, 14, ..... is 109?
Sol. Let 109 is the nth term,
∴ Using Tn = a + (n - 1) d, we have:
109 = 4 + (n - 1) 5
[∵ a = 4 and d = 9 - 4 = 5]

⇒ n = 21 + 1 = 22
Thus, the 22nd term is 109.

Q4. If a, (a - 2) and 3a are in A.P. then what is the value of a?
Sol. ∵ a, (a - 2) and 3a are in A.P.
∴ (a - 2) - a = 3a - (a - 2)
⇒ a - 2 - a = 3a - a + 2
⇒ - 2 = 2a + 2
⇒ 2a = - 2 - 2 = - 4
⇒ a = -4/2 = -2
Thus, the required value of a is - 2.

Q5. How many terms are there in the A.P.?
7, 10, 13, ....., 151
Sol. Here, a = 7,  d = 10 - 7 = 3
Let there are n-terms.
∴ Tn = a + (n - 1) d
⇒ T51 = 7 + (n - 1) × 3

⇒ 144/3 = n − 1 ⇒ n = 48 + 1 = 49
i.e., n = 49

Q6. Which term of the A.P. 72, 63, 54, ..... is 0?
Sol. Here, a = 72
d = 63 - 72 = - 9
Let nth term of this A.P. be 0
∴ Tn = a + (n - 1) d
⇒ 72 + (n - 1) × (- 9) = 0
⇒ (n - 1) = -72/-9 = 8
⇒ n = 8 + 1 = 9
Thus the 9th term of the A.P. is 0.

Q7. The first term of an A.P. is 6 and its common difference is - 2. Find its 18th term.
Sol. Using Tn = a + (n - 1) d, we have:
T18 = 6 + (18 - 1) × (- 2)
= 6 + 17 × (- 2)
= 6 - 34 = - 28
Thus, the 18th term is - 28.

Q8. The 4th term of an A.P. is 14 and its 12th term 70. What is its first term
Sol. Let the first term = a
If ‘d’ is the common difference,
Then T4 = a + 3d = 14    ...(1)
And T12 = a + 11d = 70   ...(2)
Subtracting (1) from (2),
a + 11d - a - 3d = 70 - 14
⇒ 8d = 56  ⇒ d = 56/8 = 7
∴ From (1), a + 3 (7) = 14
⇒ a + 21 = 14
⇒ a = 14 - 21 = (- 7)
Thus, the first term is - 7.

Q9. Which term of A.P. 5, 2, - 1, - 4 ..... is - 40?
Sol. Here, a = 5
d = 2 - 5 = - 3
Let nth term be - 40
∴ Tn = a + (n - 1) d
⇒ - 40 = 5 + (n - 1) × (- 3)

⇒ n = 15 + 1 = 16 i.e.,
i.e., The 16th term of the A.P. is - 40.

Q10. What is the sum of all the natural numbers from 1 to 100?
Sol.

We have:
1, 2, 3, 4, ....., 100 are in an A.P. such that
a = 1  and  l = 100
∴ Sn = n/2 [a + l]
⇒ S100 =100/2 [1 + 100]
= 50 × 101 = 5050.

Q11. For an A.P., the 8th term is 17 and the 14th term is 29. Find its common difference.
Sol. Let the common difference = d and first term = a
∴ T8 = a + 7d = 17 ...(1)
T14 = a + 13d = 29 ...(2)
Subtracting (1) from (2), we have:
a + 13d - a - 7d = 29 - 17
⇒ 6d = 12
⇒ d = 12/6 = 2
∴ The required common difference = 2.

Q12. If the first and last terms of an A.P. are 10 and - 10. How many terms are there? Given that d = - 1.
Sol. Let the required number of terms is n and
1st term a = 10
nth term Tn = - 10
Let common difference be d then using,
Tn = a + (n - 1) d, we have:
- 10 = 10 + (n - 1) × (- 1)
⇒ - 10 = 10 - n + 1
⇒ - n + 1 = - 10 - 10 = - 20
⇒ - n = - 20 - 1 = - 21
⇒ n = 21

13. The nth term of an A.P. is (3n - 2) find its first term.
Sol.
∵ Tn = 3n - 2
∴ T1 = 3 (1) - 2 = 3 - 2 = 1
⇒   First term = 1

Q14. The nth term of an A.P. is (2n - 3) find the common difference.
Sol. Here, Tn = 2n - 3
∴ T1 = 2 (1) - 3 = - 1
T2 = 2 (2) - 3 = 1
∴ d = T2 - T1 = 1 - (- 1) = 2
Thus the common difference is 2.

Q15. If the nth term of an A.P. is (7n - 5). Find its 100th term.
Sol.
Here, Tn = 7n - 5
∴ T1 = 7 (1) - 5 = 2
T2 = 7 (2) - 5 = 9
∴ a = 2
and d = T2 - T1
= 9 - 2 = 7
Now T100 = 2 + (100 - 1) 7
[using Tn = a + (n - 1) d]
= 2 + 99 × 7
= 2 + 693 = 695.

Q16. Find the sum of first 12 terms of the A.P. 5, 8, 11, 14, ...... .
Sol. Here, a = 5
d = 8 - 5 = 3
n = 12
Using Sn = n/2 [2 (a) + (n - 1) d]
we have: S12 = 12/2 [2 (5) + (12 - 1) × 3]
= 6 [10 + 33]
= 6 × 43 = 258

Q17. Write the common difference of an A.P. whose nth term is 3n + 5.
Sol. Tn = 3n + 5
∴ T1 = 3 (1) + 5 = 8
T2 = 3 (2) + 5 = 11
⇒ d = T2 - T1
= 11 - 8 = 3
Thus, the common difference = 3.

Q18. Write the value of x for which x + 2, 2x, 2x + 3 are three consecutive terms of an A.P.
Sol.
Here,
T1 = x + 2
T2 = 2x
T3 = 2x + 3
For an A.P., we have:
∴ 2x - (x + 2) = 2x + 3 - 2x
⇒ 2x - x - 2 = 2x + 3 - 2x
⇒ x - 2 = 3
⇒ x = 3 + 2 = 5
Thus, x = 5

Q19. What is the common difference of an A.P. whose nth term is 3 + 5n?
Sol. ∵ Tn = 3 + 5n
∴ T1 = 3 + 5 (1) = 8
And T2 = 3 + 5 (2) = 13
∵ d = T2 - T1
∴ d = 13 - 8 = 5
Thus, common difference = 5.

Q20. For what value of k, are the numbers x, (2x + k) and (3x + 6) three consecutive terms of an A.P.?
Sol. Here, T1 = x,  T2 = (2x + k)  and  T3 = (3x + 6)
For an A.P., we have
T2 - T1 = T3 - T2
i.e., 2x + k - x = 3x + 6 - (2x + k)
⇒ x + k = 3x + 6 - 2x - k
⇒ x + k = x + 6 - k
⇒ k + k = x + 6 - x
⇒ 2k = 6
⇒ k = 6/2 = 3

Q21. For what value of k, will k + 9, 2k - 1 and 2k + 7 are consecutive terms of an A.P.?
Sol. ∵ T1 = k + 9,   T2 = 2k - 1  and   T3 = 2k + 7
For an A.P., T2 - T1 = T3 - T2
∴   (2k - 1) - (k + 9) = (2k + 7) - (2k - 1)
⇒ 2k - 1 - k - 9 = 2k + 7 - 2k + 1
⇒ k - 10 = 8
⇒ k   = 8 + 10
⇒ k   = 18

Q22. If 4/5, a, 2 are three consecutive terms of an A.P., then find the value of a?
Sol. Here,
T1 = 4/5
T2 = a
T3 = 2
∵ For an A.P.,
T2 - T1 = T3 - T2

⇒ 2a = 14/5

Thus, a = 7/5

Q23. For what value of p are 2p - 1, 7 and 3p three consecutive terms of an A.P.?
Sol. Here,
T= 2p - 1
T2 = 7
T3 = 3p
∵ For an A.P., we have:
T2 - T1 = T3 - T2
⇒ 7 - (2p - 1) = 3p - 7
⇒ 7 - 2p + 1 = 3p - 7
⇒ - 2p - 3p = - 7 - 1 - 7
⇒ - 5p = - 15
⇒ p = -15/-5 = 3
Thus, p = 3

Q24. For what value of p are 2p + 1, 13 and 5p - 3 three consecutive terms of an A.P.?
Sol. Here,
T= 2p + 1
T2 = 13
T3 = 5p - 3
For an A.P., we have:
T2 - T1 = T3 - T2
⇒ 13 - (2p + 1) = 5p - 3 - 13
⇒ 13 - 2p - 1 = 5p - 16
⇒ - 2p + 12 = 5p - 16
⇒ - 2p - 5p = - 16 - 12 = - 28
⇒ - 7p = - 28

∴ p = 4

Q25. The nth term of an A.P. is 7 - 4n. Find its common difference.
Sol. ∵ Tn = 7 - 4n
∴ T1 = 7 - 4 (1) = 3
T2 = 7 - 4 (2) = - 1
∴ d = T2 - T1
= (- 1) - 3 = - 4
Thus, common difference = - 4

Q26. The nth term of an A.P. is 6n + 2. Find the common difference.
Sol.
Here, Tn = 6n + 2
∴ T1 = 6 (1) + 2 = 8
T2 = 6 (2) + 2 = 14
⇒ d = T2 - T1 = 14 - 8 = 6
∴ Common difference = 6.

Q27. Write the next term of the A.P.  √8, √18 √32 ......
Sol
.
Here,

Now, d = T2 − T1

∴ T4 = a + 3d

Thus, the next term of the A.P. is  5√2 or

Q28. The first term of an A.P. is p and its common difference is q. Find the 10th term.
Sol. Here, a = p  and  d = q
∵ Tn = a + (n - 1) d
∴ T10 = p + (10 - 1) q
= p + 9q
Thus, the 10th term is p + 9q.

Q29. (a) Find the next term of the A.P.
(b) Find the tenth term of the sequence
Sol.
(a) Here,

Now, d = T2 − T1

Now, using Tn = a + (n - 1) × d, we have
T4 = a + 3d

Thus, the next term = √32.
(b) Here, T10 = a + 9d

Q30. Which term of the A.P.:
21, 18, 15, ..... is zero?
Sol. Here, a = 21
d = 18 - 21 = - 3
Since Tn = a + (n - 1) d
⇒ 0 = 21 + (n - 1) × (- 3)
⇒ - 3 (n - 1) = - 21
⇒ (n - 1) = -21/-3 = 7
⇒ n = 7 + 1 = 8
Thus, the 8th term of this A.P. will be 0.

Q31. Which term of the A.P.:
14, 11, 8, ..... is - 1?
Sol. Here, a = 14
d = 11 - 14 = - 3
Let the nth term be (- 1)
∴ Using Tn = a + (n - 1) d, we get
- 1 = 11 + (n - 1) × (- 3)
⇒ - 1 - 14 = - 3 (n - 1)
⇒ - 15 = - 3 (n - 1)
∴ n - 1 = -15/-3 = 5
⇒ n = 5 + 1 = 6
Thus, -1 is the 6th term of the A.P.

Q32. The value of the middlemost term (s) of the AP : –11, –7, –3, ...49.
Sol. ∵ a = –11,  an = 49  and   d = (–7) – (–11) = 4
∴ an = a + (n – 1)d
⇒ 49 = –11 + (n – 1) × 4
⇒   n = 16
Since, n is an even number
∴ There will be two middle terms, which are:

or   8th   and 9th
Now, a8 = a + (8 – 1)d
= –11 + 7 × 4 = 17
a9 = a + (9 – 1)d
= –11 + 8 × 4 = 21
Thus, the values of the two middlemost terms are 17 and 21.

The document Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions

 1. What is an arithmetic progression?
Ans. An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is always the same. This difference is called the common difference. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
 2. How is the nth term of an arithmetic progression calculated?
Ans. The nth term of an arithmetic progression can be calculated using the formula: nth term = first term + (n - 1) × common difference For example, if the first term is 3 and the common difference is 2, the 5th term would be: 5th term = 3 + (5 - 1) × 2 = 3 + 4 × 2 = 11.
 3. Can an arithmetic progression have negative terms?
Ans. Yes, an arithmetic progression can have negative terms. The common difference can be positive, negative, or zero. For example, -4, -1, 2, 5, 8 is an arithmetic progression with a common difference of 3.
 4. How can we find the sum of the first n terms of an arithmetic progression?
Ans. The sum of the first n terms of an arithmetic progression can be calculated using the formula: Sum = (n/2) × (first term + last term) Alternatively, the sum can also be calculated using the formula: Sum = (n/2) × (2 × first term + (n - 1) × common difference)
 5. Can the common difference of an arithmetic progression be zero?
Ans. Yes, the common difference of an arithmetic progression can be zero. In this case, all the terms of the progression would be the same. For example, 2, 2, 2, 2, 2 is an arithmetic progression with a common difference of zero.

## Mathematics (Maths) Class 10

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