Mathematics
Multiple Choice Questions (MCQ)
MCQ | Class 10 | Chapter 5
ARITHMATIC PROGRESSION
- MCQ Based on the general Arithmetic Progression.
- MCQ Based on the nth term of Arithmetic Progression.
- MCQ Based on the sum of n terms of Arithmetic Progression.
- MCQ Based on the word problems of Arithmetic Progression.
- MCQ Based from the nth term from the end of the sequence.
Features
- In this pdf given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.
- Solution Hints are also given to some difficult problems.
- Each MCQ contains four options from which one option is correct.
- On the right hand side column of the pdf Answer option is given.
Action Plan
- First of all students should Learn and write all basic points and Formulas related to the Arithmetic Progression.
- Start solving the NCERT Problems with examples.
- Solve the important assignments on the Arithmetic Progression.
- Then start solving the following MCQ.
MCQ | CHAPTER 5 | CLASS 10
Q 1) In an Arithmetic Progression, if a = 28, d = – 4, n = 7, then an is:
a) 4
b) 5
c) 3
d) 7
a) 10, 30, 50, 60
b) 10, 20, 30, 40
c) 10, 15, 20, 25
d) 10, 18, 20, 30
Ans: b
a) 97
b) 77
c) – 77
d) – 87
Ans: c
a) 28
b) 22
c) – 38
d) – 48
a) 11 and 9
b) 17 and 9
c) 18 and 8
d) 18 and 9
a) 12th
b) 13th
c) 15th
d) 16th
a) 17
b) 137
c) 143
d) -143
a) 50
b) 40
c) 60
d) 30
a) 147
b) 151
c) 154
d) 158
a) 45
b) 55
c) 65
d) 75
a) 6
b) 7
c) 20
d) 28
a) an AP with d = –16
b) an AP with d = 4
c) an AP with d = –4
d) not an AP
a) 30
b) 33
c) 37
d) 38
a) 9th
b) 10th
c) 11th
d) 12th
a) 8
b) – 8
c) – 4
d) 4
a) Pythagoras
b) Newton
c) Gauss
d) Euclid
a) –320
b) 320
c) –352
d) –400
a) 7
b) 3
c) 4
d) 1
a) 2, 4, 6
b) 1, 5, 3
c) 2, 8, 4
d) 2, 3, 4
Explanation: Reason: Let three numbers be a – d, a, a + d
∴ a – d + a + a + d = 9
⇒ 3a = 9 ⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2
a) – 955
b) – 945
c) – 950
d) – 965
Q 21) The sum of all two digit odd numbers is
a) 2575
b) 2475
c) 2524
d) 2425
(a) 262
(b) 272
(c) 282
(d) 292
(a) 2n²
(b) 2n + 1
(c) 2n – 1
(d) n²
(a) na
(b) (2n – 1) a
(c) n²a
(d) n²a²
(a) 0
(b) 2
(c) 4
(d) 6
(a) 8th
(b) 10th
(c) 9th
(d) 11th
(a) 18
(b) 9
(c) 77
(d) 0
Q 28) If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
(a) 2√7
(6) 6√2
(c) 9√2
(d) 7√2
(a) s – p
(b) s – q
(c) s – r
(d) none of these
Q 31) If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
(a) 2, 4, 6
(b) 1, 5, 3
(c) 2, 8, 4
(d) 2, 3, 4
Q 32) If the sum of the first m terms of an AP is n and the sum of its n terms is m, then the sum of its (m + n) terms will be
(a) m + n
(b) -(m + n)
(c) m – n
(d) 0
(a) 5n + 2
(b) 5n + 3
(c) 5n – 5
(d) 5n – 3
(a) 0
(b) 1
(c) -1.
(d) 2
Ans: a
∴ b = a + d, c = a + 2d, d = a + 3d and e = a + 4d
Given equation a-4b + 6c-4d + c
= a – 4(a + d) + 6(a + 2d) – 4(a + 3d) + (a + 4d)
= a – 4a – 4d + 6a + 12d – 4a – 12d + a + 4d = 8a – 8a + 16d – 16d = 0
(a) a + nd
(b) a – (n – 1)d
(c) a + (n – 1)d
(d) n + nd
(a) 0
(b) 2
(c) 4
(d) 6
(a) 209
(b) 205
(c) 214
(d) 213
(a) A + B
(b) A – B
(c) 2A
(d) 2B
(a) 17
(b) 867
(c) 876
(d) 786
(a) 920
(b) 860
(c) 900
(d) 960
(a) 17
(b) 137
(c) 143
(d) –143
Q 42) If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37