# Free Arithmetic Progressions 01 Practice Test - 10th Grade

### Question 1

If x+1,3x and 4x+2 are the first three terms of an AP, then its 5th term is ___.

14

19

24

28

#### SOLUTION

Solution :C

Given, x+1,3x,4x+2 are in AP.

So, the difference of any two consecutive terms will be the same.

∴3x–(x+1)=(4x+2)–3x

⇒ 2x−1=x+2

⇒ x=3So, the first term of the AP is a=x+1=3+1=4.

The second term of the AP is 3x=3×3=9.

∴Common difference, d=9–4=5The nth term of an AP with first term a and common difference d is given by

tn=a+(n−1)d.

⇒t5=4+4(5)=24

### Question 2

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

150

178

210

185

#### SOLUTION

Solution :B

Given that

a11=38 and a16=73,

where a11 is the 11th term and a16 is the 16th term of an AP.

The nth term of an AP with first term a and common difference d is given by

an=a+(n−1)d.

⇒38=a+10d ....(i)

73=a+15d ....(ii)

Subtracting equation (i) from equation (ii), we get

35=5d.

⇒d=7

Substituting the value of d in equation (i), we get

a=38−10d=38−10(7)=−32.

Now, a31=a+(31−1)d =−32+30(7)

=178

Therefore, the 31st term of the given AP is 178.

### Question 3

Find the number of all natural numbers that lie between 24 and 101, which are divisible by 5.

15

16

18

20

#### SOLUTION

Solution :B

Natural numbers that lie between 24 and 101,which are divisible by 5 are 25, 30,... and 100.

This is an A.P with first term a=25, common difference d=5 and the last term l=100.

But, l=a+(n−1)d.

⇒100=25+(n−1)×5

⇒ 75=(n−1)×5

⇒ 15=(n−1)

⇒ 16=n

Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.

### Question 4

If the third and the ninth terms of an AP are 4 and -8 respectively, which term of this AP is zero?

5th term

4th term

3rd term

6th term

#### SOLUTION

Solution :A

Given, a3=4 and a9=−8, where a3 and a9 are the third and ninth terms of an AP respectively.

Using the formula for nth term of an AP with first term a and common difference d,

an=a+(n−1)d, we get

a3=a+(3−1)d and a9 = a+(9−1)d.

⇒4 = a+2d...(i)

−8= a+8d...(ii)

Substituting the value of a from equation (i) in equation (ii), we have

−8=4−2d+8d

⇒−12 = 6d

⇒ d =−126 =−2

Solving for a, we get −8 = a−16.

⇒a = 8

Therefore, the first term of the AP is 8 and common difference is −2.

Let the nth term of the AP be zero.

i.e., an=0

a+(n−1)d=0

8+(n−1)(−2)=0

⇒ 8−2n+2=0

⇒ 2n=10

⇒ n=102=5

Therefore, the 5th term of the AP is equal to 0.

### Question 5

If the 7th and 13th terms of an AP are 34 and 64, then its 18th term is ___.

#### SOLUTION

Solution :C

The nth term of an AP of first term a and common difference d is

an=a+(n−1)d.⇒a7=a+6d=34−−−(1) a13=a+12d=64−−−(2)(2) − (1)⇒6d=30 ⇒d=5Substituting d = 5 in (1), we get a=4.∴a18=a+17d=4+(17)5=89

### Question 6

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

#### SOLUTION

Solution :Here, S14 = 1050, n = 14, a = 10.

As Sn = (n2) [2a + (n-1) d]

So,

1050=(142) [20 + 13d]

⇒1050=140+91d

⇒910=91d

⇒d=10

Therefore, a20=10+(20–1)×10=200

i.e. 20th term is 200.

### Question 7

The sum of r terms of an AP is 2r2+3r. The nth term is ___.

2n−1

4n+1

3n−1

3n+1

#### SOLUTION

Solution :B

Given, Sr=(2r2+3r), the sum upto r terms of an AP.

We know that the nth term of AP is

Tn=Sn−Sn−1.

⇒Sn=2n2+3n and

Sn−1=2(n−1)2+3(n−1).

∴Tn=Sn−Sn−1

=2n2+3n−[2(n2−2n+1)+3n−3]

=2n2+3n−2n2+4n−2−3n+3 =4n+1

### Question 8

Find the difference between the sum of first 100 natural numbers and the sum of even numbers from 1 to 100.

2200

2400

2500

2700

#### SOLUTION

Solution :C

Using the formula

Sn=n2[2n+(n−1)d]

Sum of first 100 natural numbers = 1002(2+99)=5050Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100.

So, we have

Sum of even numbers from 1 to 100 = 502(2×2+49×2) = 2550Hence, the difference is 5050 - 2550 = 2500

### Question 9

Find the sum of the first 24 terms of an AP whose nth term is given by tn=3+2n.

568

624

564

672

#### SOLUTION

Solution :D

Given, tn=3+2n.

∴t1=3+2×1=5t2=3+2×2=7

t3=3+2×3=9

Note that t2−t1=t3−t2=2.

Thus, the common difference of the AP is 2.

Therefore, the AP is 5, 7, 9, 11 . . .

The sum to n terms of an AP with first term a and common difference d is

Sn=n2[2a+(n−1)d].∴S24=242(2(5)+(24−1)2)

=12(10+46)=672

### Question 10

Find the sum of the first 15 multiples of 8.

1000

960

500

820

#### SOLUTION

Solution :B

The first 15 multiples of 8 are 8, 16, 24, 32, .... 112, 120

First term, a=8

Common difference, d=16−8=8

n=15

Applying formula, Sn=n2(2a+(n−1)d) to find sum of n terms of AP, we get

S15=152(2(8)+(15−1)(8))

=152(16+8(14))

=152(16+112)

=152(128)=960