Question

Can anyone give me set of MCQs of Arithmetic Progressions class 10? ASAP
Please help me tomorrow is my test.


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Answer 1

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

Answer 2

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