Can anyone give me set of MCQs of Arithmetic Progressions class 10? ASAP
Please help me tomorrow is my test.
Answers
Answer:
Step-by-step explanation:
1. The nth term of an A.P. is given by an = 3 + 4n. The common difference is
(a) 7
(b) 3
(c) 4
(d) 1
Answer/Explanation
2. If p, q, r and s are in A.P. then r – q is
(a) s – p
(b) s – q
(c) s – r
(d) none of these
Answer/Explanation
3. 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
Answer/Explanation
4. The (n – 1)th term of an A.P. is given by 7,12,17, 22,… is
(a) 5n + 2
(b) 5n + 3
(c) 5n – 5
(d) 5n – 3
Answer/Explanation
5. The nth term of an A.P. 5, 2, -1, -4, -7 … is
(a) 2n + 5
(b) 2n – 5
(c) 8 – 3n
(d) 3n – 8
Answer/Explanation
6. The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is
(a) -955
(b) -945
(c) -950
(d) -965
Answer/Explanation
7. Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4
(a) 262
(b) 272
(c) 282
(d) 292
Answer/Explanation
8. The sum of all two digit odd numbers is
(a) 2575
(b) 2475
(c) 2524
(d) 2425
Answer/Explanation
9. The sum of first n odd natural numbers is
(a) 2n²
(b) 2n + 1
(c) 2n – 1
(d) n²
Answer/Explanation
10. If (p + q)th term of an A.P. is m and (p – q)tn term is n, then pth term is
MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers 1
Answer/Explanation
Step-by-step explanation:
Below are the MCQs for Arithmetic Progression
1.In an Arithmetic Progression, if a=28, d=-4, n=7, then an is:
(a)4
(b)5
(c)3
(d)7
Answer: a
Explanation: For an AP,
an = a+(n-1)d
= 28+(7-1)(-4)
= 28+6(-4)
= 28-24
an=4
2.If a=10 and d=10, then first four terms will be:
(a)10,30,50,60
(b)10,20,30,40
(c)10,15,20,25
(d)10,18,20,30
Answer: b
Explanation: a = 10, d = 10
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
3.The first term and common difference for the A.P. 3,1,-1,-3 is:
(a)1 and 3
(b)-1 and 3
(c)3 and -2
(d)2 and 3
Answer: c
Explanation: First term, a = 3
Common difference, d = Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2
4.30th term of the A.P: 10,7, 4, …, is
(a)97
(b)77
(c)-77
(d)-87
Answer: c
Explanation: Given,
A.P. = 10, 7, 4, …
First term, a = 10
Common difference, d = a2 − a1 = 7−10 = −3
As we know, for an A.P.,
an = a +(n−1)d
Putting the values;
a30 = 10+(30−1)(−3)
a30 = 10+(29)(−3)
a30 = 10−87 = −77
5.11th term of the A.P. -3, -1/2, ,2 …. Is
(a)28
(b)22
(c)-38
(d)-48
Answer: b
Explanation: A.P. = -3, -1/2, ,2 …
First term a = – 3
Common difference, d = a2 − a1 = (-1/2) -(-3)
⇒(-1/2) + 3 = 5/2
Nth term;
an = a+(n−1)d
Putting the values;
a11 = 3+(11-1)(5/2)
a11 = 3+(10)(5/2)
a11 = -3+25
a11 = 22
6.The missing terms in AP: __, 13, __, 3 are:
(a)11 and 9
(b)17 and 9
(c)18 and 8
(d)18 and 9
Answer: (c)
Explanation: a2 = 13 and
a4 = 3
The nth term of an AP;
an = a+(n−1) d
a2 = a +(2-1)d
13 = a+d ………………. (i)
a4 = a+(4-1)d
3 = a+3d ………….. (ii)
Subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
Now put value of d in equation 1
13 = a+(-5)
a = 18 (first term)
a3 = 18+(3-1)(-5)
= 18+2(-5) = 18-10 = 8 (third term).
7. Which term of the A.P. 3, 8, 13, 18, … is 78?
(a)12th
(b)13th
(c)15th
(d)16th
Answer: (d)
Explanation: Given, 3, 8, 13, 18, … is the AP.
First term, a = 3
Common difference, d = a2 − a1 = 8 − 3 = 5
Let the nth term of given A.P. be 78. Now as we know,
an = a+(n−1)d
Therefore,
78 = 3+(n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
8.The 21st term of AP whose first two terms are -3 and 4 is:
(a)17
(b)137
(c)143
(d)-143
Answer: b
Explanation: First term = -3 and second term = 4
a = -3
d = 4-a = 4-(-3) = 7
a21=a+(21-1)d
=-3+(20)7
=-3+140
=137
9. If 17th term of an A.P. exceeds its 10th term by 7. The common difference is:
(a)1
(b)2
(c)3
(d)4
Answer: (a)
Explanation: Nth term in AP is:
an = a+(n-1)d
a17 = a+(17−1)d
a17 = a +16d
In the same way,
a10 = a+9d
Given,
a17 − a10 = 7
Therefore,
(a +16d)−(a+9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.
10. The number of multiples of 4 between 10 and 250 is:
(a)50
(b)40
(c)60
(d)30
Answer: (c)
Explanation: The multiples of 4 after 10 are:
12, 16, 20, 24, …
So here, a = 12 and d = 4
Now, 250/4 gives remainder 2. Hence, 250 – 2 = 248 is divisible by 2.
12, 16, 20, 24, …, 248
So, nth term, an = 248
As we know,
an = a+(n−1)d
248 = 12+(n-1)×4
236/4 = n-1
59 = n-1
n = 60
11. 20th term from the last term of the A.P. 3, 8, 13, …, 253 is:
(a)147
(b)151
(c)154
(d)158
Answer: (d)
Explanation: Given, A.P. is 3, 8, 13, …, 253
Common difference, d= 5.
In reverse order,
253, 248, 243, …, 13, 8, 5
So,
a = 253
d = 248 − 253 = −5
n = 20
By nth term formula,
a20 = a+(20−1)d
a20 = 253+(19)(−5)
a20 = 253−95
a20 = 158
12. The sum of the first five multiples of 3 is:
(a)45
(b)55
(c)65
(d)75
Answer: (a)
Explanation: The first five multiples of 3 is 3, 6, 9, 12 and 15
a=3 and d=3
n=5
Sum, Sn = n/2[2a+(n-1)d]
S5 = 5/2[2(3)+(5-1)3]
=5/2[6+12]
=5/2[18]
=5 x 9
= 45