# Free Arithmetic Progressions 03 Practice Test - 10th Grade

### Question 1

Find the sum of the following AP 1, 3, 5, 7 …199.

#### SOLUTION

Solution :A

Given,a=1,d=2,an=l=199a+(n−1)d=199

1+(n−1)2=199

⇒1+2n−2=199

⇒2n=200

n=100

Sn=n2(a+l)

Sn=1002(1+199)

Sn=10000

### Question 2

The sum of four consecutive numbers in an A.P. with common difference d>0 is 20. If the sum of their squares is 120, then the middle terms are __ and __.

2,4

4,6

6,8

8,10

#### SOLUTION

Solution :B

Let the numbers be a−3d,a−d,a+d and a+3d. These numbers form an AP with first term a−3d and common difference 2d>0.

Also, given that (a−3d)+(a−d)+(a+d)+(a+3d)=20.

⇒4a=20

⇒a=5

Also, (a−3d)2+(a−d)2+(a+d)2+(a+3d)2=120

⇒4a2+20d2=120

⇒4(5)2+20d2=120

⇒d2=1

⇒d=±1

SIince d>0, we must have d=1.

Hence, the numbers are 2, 4, 6, and 8.

### Question 3

Which term of the AP 3, 15, 27, 39, .... will be 132 more than its 54th term?

#### SOLUTION

Solution :C

a=3

d=15−3=12

Let the required term be the nth term.

an=a+(n−1)d

a54+132=(32+53×12)+132

=32+636+132=800=an

800=32+(n−1)12

768=(n−1)12n−1=76812=64n=64+1=65

Thus, 65th term is 132 more than 54th term.

### Question 4

Find the sum : 34 + 32 + 30 + .... + 10

300

286

310

240

#### SOLUTION

Solution :B

34+32+30+....+10

This sequence is an AP.

Here, a=34,d=32−34=−2.

Let the number of terms of the AP be n.

We know that

an=a+(n−1)d.⇒10=34+(n−1)(−2)

⇒(n−1)(−2)=−24

⇒n−1=242=12

⇒n=13

Also, we know that the sum to n terms of an AP with first term a and last term l is given by

Sn=n2(a+l).∴S13=132(34+10)

S13=286

Hence, the required sum is 286.

### Question 5

Find the sum to 100 terms of the series:

1 + 4 + 7 + 5 + 13 + 6 ...

#### SOLUTION

Solution :B

The series can be clubbed into two AP's, (1 + 7 + 13......) and (4 + 5 + 6.......). So, we can observe that the series can be taken as two AP's by simple rearrangement of numbers. Now, we can find the Sum of 50 terms for each AP formed.

Formula for sum of first n terms of AP is: Sn=n2(2a+(n−1)d)

So, for first AP

S1=502(2+49×6)=7400

and similarly,

S2=502(8+49)=1425

So, Sum of first 100 terms

=7400+1425=8825

### Question 6

If the 11th and13th terms of an AP are 35 and 41 respectively, then its common difference is ___.

#### SOLUTION

Solution :D

The nth term of an A.P, with first term a and common difference d is given by

Tn=a+(n−1)d.

⇒a11=a+10d=35...(1)

a13=a+12d=41...(2)

Solving equations (1) and (2), we get

2d=6.

⇒d=3

### Question 7

The sum of first ten terms of an A.P. is four times the sum of its first five terms. What is the ratio of first term and common difference?

1

12

4

14

#### SOLUTION

Solution :B

Let S10 be the sum of first 10 terms and S5 be the sum of first 5 terms.

The sum upto n terms of an AP of first term a and common difference d is given by

Sn=n2[2a+(n−1)d].Given, S10=4S5.

⇒102[2a+(10−1)d]=4×52[2a+(5−1)d]

⇒102[2a+9d]=4×52[2a+4d]

⇒ 2a+9d=4a+8d

⇒ 2a=d

⇒ ad=12

### Question 8

An AP consists of 50 terms of which the third term is 12 and the last term is 106. Find the 29th term.

#### SOLUTION

Solution :It is also given that a3=12 and a50=106

Using formula an=a+(n−1)d to find nth term of AP, we get

a50=a+(50−1)d⇒106=a+49d⋯(1)

a3=a+(3−1)d⇒12=a+2d⋯(2)

Subtracting (1) and (2), we get,

⇒47d=94

⇒d=2

Substituting d in (2), we get,

a+2(2)=12

a=12−4=8The 29th term is,

a29=a+(29−1)d=8+28(2)=8+56=64

### Question 9

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

#### SOLUTION

Solution :It is given that 17th term exceeds its 10th term by 7

It means a17 = a10 + 7......(1)

Here, a17 is the 17th term and a_{10}is the 10th term of an AP.

Using formula an=a+(n−1)d to find nth term of arithmetic progression, we geta17 =a+(16)d .....(2)

a10 = a+(9)d .....(3)

Putting (2) and (3) in equation (1), we get

a+16d=a+9d+7

⇒7d=7

⇒d=77=1

### Question 10

In the arithmetic progression given below, find the common difference.

1, a, b, c, 9

0

3

1

2

#### SOLUTION

Solution :D

There are five terms in this AP

an = a1 + (n - 1)d

⇒ an−a1=(n−1)d

⇒ an−a1n−1=dHere, a1=1; n=5; an=9

Therefore the common difference,

d=9−15−1=2