This set of Class 10 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Arithmetic Progressions”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. A sequence in which each term differs from its preceding term by a constant is called arithmetic progression.
a) True
b) False
View Answer
Explanation: Consider an example, 8, 11, 14, 17, 20, ….
Now, 11 – 8 = 3
14 – 11 = 3
Here the constant is 3. This constant is also termed as common difference.
2. If we add the terms of an AP, then we get a corresponding series.
a) True
b) False
View Answer
Explanation: For example, consider an AP 2, 11, 20, 29……
Adding the terms of the AP we get the arithmetic series as 2 + 11 + 20 + 29 + ⋯
3. Sequences which follow a definite pattern called progressions.
a) True
b) False
View Answer
Explanation: Let us consider an example,
2, 4, 6, 8, …
Here each term is twice the preceding term.
Since it follows a definite pattern it is called as progressions.
Progressions are of three types: Arithmetic progression, Geometric progression and Harmonic progression.
4. If the first term of an AP is a and its common difference is d then the nth term is given by Tn = a + (n – 1)d
a) True
b) False
View Answer
Explanation: In the given AP we have, first term as a and common difference as d.
So, the given AP may be written as a, a + d, a + 2d, a + 3d, …
In this AP, we have,
First term = a = a + (1 – 1)d
Second term = a = a + (2 – 1)d
Third term = a = a + (3 – 1)d
And so, on
Hence, the nth term of the AP will be a + (n – 1)d.
5. Which term of the AP 2, 11, 20, 29…… is 290?
a) 32
b) 35
c) 30
d) 33
View Answer
Explanation: Here a = 2 and d = 11 – 2 = 9
Let the given AP contain n terms.
Tn = 290
a + (n – 1)d = 290
2 + (n – 1)9 = 290
2 + 9n – 9 = 290
9n = 297
n = 33
Thus, the 33 term of the AP is 290.
6. How many terms are there in the AP 9, 16, 23, 30……………282?
a) 50
b) 45
c) 40
d) 42
View Answer
Explanation: Here a = 9 and d = 16 – 9 = 7
Let the given AP contain n terms.
Tn = 282
a + (n – 1)d = 282
9 + (n – 1)7 = 282
9 + 7n – 7 = 282
7n = 280
n = 40
Thus, the given AP has 40 terms.
7. The middle term of the AP 11, 8, 5, … , -79 will be?
a) 30
b) -34
c) 34
d) -30
View Answer
Explanation: Here a = 11 and d = 8 – 11 = -3
Let the given AP contain n terms.
Tn = -79
a + (n – 1)d = -79
11 + (n – 1) – 3 = -79
11 – 3n + 3 = -79
14 + 79 = 3n
n = 31
Thus, the given AP has 31 terms.
∴ its middle term = \(\frac {1}{2}\) (31 + 1) = 16th term
∴ 16th term = 11 + (16 – 1) – 3 = 11 – 45 = -34
8. Which term of the AP 80, 76, 72, 68……. is the first negative term?
a) 22
b) 21
c) 23
d) 24
View Answer
Explanation: Here a = 80 and d = 76 – 80 = – 4
Let Tn be the first negative term of the given AP.
Tn < 0
a + (n – 1)d < 0
80 + (n – 1) – 4 < 0
80 – 4n + 4 < 0
84 < 4n
n > 21
Hence the 22th term of the given AP will be its first negative term.
9. What is the arithmetic mean between a2 and b2 ?
a) \(\frac {1}{2}\) (a – b)
b) \(\frac {1}{2}\) (a + b)
c) \(\frac {1}{2}\) (a2 – b2)
d) \(\frac {1}{2}\) (a2 + b2)
View Answer
Explanation: We know that the arithmetic mean between two numbers is \(\frac {1}{2}\) (a + b)
∴ Arithmetic mean between a2 and b2 = \(\frac {1}{2}\) (a2 + b2)
10. What will be the value of k so that (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP?
a) 4
b) 3
c) \(\frac {3}{4}\)
d) \(\frac {4}{3}\)
View Answer
Explanation: Since, (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP.
The common difference will be same.
(5k – 3) – (2k + 1) = (-8k + 5) – (5k – 3)
3k – 4 = -13k + 8
16k = 12
k = \(\frac {3}{4}\)
11. The sum of three numbers is 30 and their product 750. What are the three numbers?
a) -2, -6 and -10
b) 2, 6, and 10
c) 5, 10 and 15
d) -5, -10 and -15
View Answer
Explanation: Let the three numbers in AP be (a + d), a, and (a – d).
Sum of the five numbers is 30.
∴ (a + d) + a + (a – d) = 30
3a = 30
a = 10
Now, product of the numbers is 750.
∴ (a + d)a(a – d) = 750
a3 – d2 a = 750
Substituting a = 10, we get
103 – d2 (10) = 750
1000 – d2 (10) = 750
250 = 10d2
d2 = 25
d = 5, -5
The three numbers are 5, 10 and 15.
12. Four numbers whose sum is 40 and sum of their squares is 480. What are the four numbers?
a) 4, 8, 12, and -16
b) 4, 8, 12, and 16
c) -4, 8, 12, and 16
d) 4, -8, 12, and 16
View Answer
Explanation: Let the four numbers be (a + 3d), (a + d), (a – 3d), (a – d).
Sum of the four numbers is 40.
a + 3d + a + d + a – 3d + a – d = 40
4a = 40
a = 10
Sum of their squares is 480.
(a + 3d)2 + (a + d)2 + (a – 3d)2 + (a – d)2 = 480
4a2 + 20d2 = 480
102 + 5d2 = 120
100 + 5d2 = 120
5d2 = 20
d2 = 4
d = 2, -2
Hence the four numbers are 4, 8, 12, and 16.
13. The sum of 5 numbers in AP is 10 and their product is 80. What are the five numbers?
a) 2 + 2√6, 2 + √6, 2, 2 – 2√6, 2 – √6
b) 2 + 2√6, 2 + √6, 2 – 2√6, 2 – √6
c) 2 + 2√6, 2 – √6, 2, 2 – 2√6, 2 – √6
d) 2 + 2√6, 2, 2 – 2√6, 2 – √6
View Answer
Explanation: Let the five numbers in AP be (a + 2d), (a + d), a, (a – 2d) and (a – d).
Sum of the five numbers is 10.
∴ (a + 2d) + (a + d) + a + (a – 2d) + (a – d) = 10
5a = 10
a = 2
Now, product of the numbers is 80.
∴ (a + 2d)(a + d)a(a – 2d)(a – d) = 80
a5 – 5a3 d2 + 4d4 a = 80
Substituting a = 2, we get
25 – 5(23 d2) + 4d4 (2) = 80
32 – 40d2 + 8d4 = 80
8d4 – 40d2 – 48 = 0
Substitute d2 = t
t2 – 5t – 6 = 0
Solving the equation, we get,
t = 6, -1
Now, t = d2
Hence, d = ±√6, ±√-1
The five numbers are 2 + 2√6, 2 + √6, 2, 2 – 2√6, 2 – √6 or 2 – 2√6, 2 – √6, 2, 2 + 2√6, 2 + √6 or 2 + 2√-1, 2 + √-1, 2, 2 – 2√-1, 2 – √-1 or 2 – 2√-1, 2 – √-1, 2, 2 + 2√-1, 2 + √-1.
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