# Free Arithmetic Progressions 01 Practice Test - 10th Grade

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

A.

14

B.

19

C.

24

D.

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
2x1=x+2
x=3

So, 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=94=5

The nth term of an AP with first term a and common difference d is given by
tn=a+(n1)d.
t5=4+4(5)=24

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

A.

150

B.

178

C.

210

D.

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+(n1)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=3810d=3810(7)=32.

Now, a31=a+(311)d     =32+30(7)
=178
Therefore, the 31st term of the given AP is 178.

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

A.

15

B.

16

C.

18

D.

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+(n1)d.
100=25+(n1)×5
75=(n1)×5
15=(n1)
16=n
Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.

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

A.

5th term

B.

4th term

C.

3rd term

D.

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+(n1)d, we get
a3=a+(31)d and a9 = a+(91)d.
4 = a+2d...(i)
8= a+8d...(ii)

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

8=42d+8d
12 = 6d
d =126 =2

Solving for a, we get 8 = a16.
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+(n1)d=0
8+(n1)(2)=0
82n+2=0
2n=10
n=102=5
Therefore, the 5th term of the AP is equal to 0.

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

A. 87
B. 88
C. 89
D. 90

#### SOLUTION

Solution : C

The nth term of an AP of first term a and common difference d is
an=a+(n1)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

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+(201)×10=200

i.e. 20th   term is 200.

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

A.

2n1

B.

4n+1

C.

3n1

D.

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=SnSn1.
Sn=2n2+3n and
Sn1=2(n1)2+3(n1).

Tn=SnSn1
=2n2+3n[2(n22n+1)+3n3]
=2n2+3n2n2+4n23n+3        =4n+1

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

A.

2200

B.

2400

C.

2500

D.

2700

#### SOLUTION

Solution : C

Using the formula
Sn=n2[2n+(n1)d]
Sum of first 100 natural numbers = 1002(2+99)=5050

Sum 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) = 2550

Hence, the difference is 5050 - 2550 = 2500

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

A.

568

B.

624

C.

564

D.

672

#### SOLUTION

Solution : D

Given, tn=3+2n.

t1=3+2×1=5

t2=3+2×2=7

t3=3+2×3=9

Note that t2t1=t3t2=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+(n1)d].

S24=242(2(5)+(241)2)

=12(10+46)=672

Find the sum of the first 15 multiples of 8.

A.

1000

B.

960

C.

500

D.

820

#### SOLUTION

Solution : B

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

First term, a=8

Common difference, d=168=8

n=15

Applying formula, Sn=n2(2a+(n1)d) to find sum of n terms of AP, we get
S15=152(2(8)+(151)(8))
=152(16+8(14))
=152(16+112)
=152(128)=960