## CBSE Class 10 Arithmetic Progression MCQ New Pattern By Clarify Knowledge

*CBSE Class 10 Arithmetic Progression MCQ New Pattern 2022*

## OUR EBOOKS CAN HELP YOU ICSE CLASS 10 BOARD

## CODE IS EASY CAN HELP YOU IN SEMESTER 2

## CBSE Class 10 Arithmetic Progression MCQ Table

**1. In an Arithmetic Progression, if a = 28, d = -4, n = 7, then a _{n} is:**

(a) 4

(b) 5

(c) 3

(d) 7

Answer: **(a) 4**

Explanation: For an AP,

a_{n} = a+(n-1)d

= 28+(7-1)(-4)

= 28+6(-4)

= 28-24

a_{n}=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) 10, 20, 30, 40**

Explanation: a = 10, d = 10

a_{1} = a = 10

a_{2} = a_{1}+d = 10+10 = 20

a_{3} = a_{2}+d = 20+10 = 30

a_{4} = a_{3}+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) 3 and -2**

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) -77**

Explanation: Given,

A.P. = 10, 7, 4, …

First term, a = 10

Common difference, d = a_{2} − a_{1} = 7−10 = −3

As we know, for an A.P.,

a_{n} = a +(n−1)d

Putting the values;

a_{30} = 10+(30−1)(−3)

a_{30} = 10+(29)(−3)

a_{30} = 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) 22**

Explanation: A.P. = -3, -1/2, 2 …

First term a = – 3

Common difference, d = a_{2} − a_{1} = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

Nth term;

a_{n} = a+(n−1)d

a_{11} = 3+(11-1)(5/2)

a_{11} = 3+(10)(5/2)

a_{11} = -3+25

a_{11} = 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: a_{2} = 13 and

a_{4} = 3

The nth term of an AP;

a_{n} = a+(n−1) d

a_{2} = a +(2-1)d

13 = a+d ………………. (i)

a_{4} = 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)

a_{3 }= 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) (d) 16th**

Explanation: Given, 3, 8, 13, 18, … is the AP.

First term, a = 3

Common difference, d = a_{2} − a_{1} = 8 − 3 = 5

Let the nth term of given A.P. be 78. Now as we know,

a_{n} = 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) 137**

Explanation: First term = -3 and second term = 4

a = -3

d = 4-a = 4-(-3) = 7

a_{21}=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) 1**

Explanation: Nth term in AP is:

a_{n }= a+(n-1)d

a_{17} = a+(17−1)d

a_{17} = 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) 60**

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, a_{n} = 248

As we know,

a_{n} = 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) 158**

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,

a_{20} = a+(20−1)d

a_{20} = 253+(19)(−5)

a_{20} = 253−95

a_{20} = 158

**12. The sum of the first five multiples of 3 is:**

(a) 45

(b) 55

(c) 65

(d) 75

Answer:** (a) 45**

Explanation: The first five multiples of 3 is 3, 6, 9, 12 and 15

a=3 and d=3

n=5

Sum, S_{n} = n/2[2a+(n-1)d]

S_{5} = 5/2[2(3)+(5-1)3]

=5/2[6+12]

=5/2[18]

=5 x 9

= 45

**13. The 10th term of the AP: 5, 8, 11, 14, … is**

(a) 32

(b) 35

(c) 38

(d) 185

Answer:** (a) 32**

Explanation:

Given AP: 5, 8, 11, 14,….

First term = a = 5

Common difference = d = 8 – 5 = 3

nth term of an AP = a_{n} = a + (n – 1)d

Now, 10th term = a_{10} = a + (10 – 1)d

= 5 + 9(3)

= 5 + 27

= 32

**14. In an AP, if d = -4, n = 7, a _{n}**

**= 4, then a is**

(a) 6

(b) 7

(c) 20

(d) 28

Answer: (**d) 28**

Solution;

Given,

d = -4, n = 7, an = 4

We know that,

a_{n} = a + (n – 1)d

4 = a + (7 – 1)(-4)

4 = a + 6(-4)

4 = a – 24

⇒ a = 4 + 24 = 28

**15. The list of numbers –10, –6, –2, 2,… is**

(a) an AP with d = –16

(b) an AP with d = 4

(c) an AP with d = –4

(d) not an AP