# Free Arithmetic Progressions 03 Practice Test - 10th Grade

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

A. 10000
B. 12000
C. 13333
D. 20000

#### SOLUTION

Solution : A

Given,a=1,d=2,an=l=199a+(n1)d=199
1+(n1)2=199
1+2n2=199
2n=200
n=100
Sn=n2(a+l)
Sn=1002(1+199)
Sn=10000

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 __.

A.

2,4

B.

4,6

C.

6,8

D.

8,10

#### SOLUTION

Solution : B

Let the numbers be a3d,ad,a+d and a+3d. These numbers form an AP with first term a3d and common difference 2d>0.
Also, given that  (a3d)+(ad)+(a+d)+(a+3d)=20.
4a=20
a=5
Also, (a3d)2+(ad)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.

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

A. 64th
B. 56th
C. 65th
D. 67th

#### SOLUTION

Solution : C

a=3
d=153=12
Let the required term be the ​nth term.
an=a+(n1)d
a54+132=(32+53×12)+132
=32+636+132=800=an
800=32+(n1)12
768=(n1)12n1=76812=64n=64+1=65
Thus, 65th term is 132 more than 54th term.

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

A.

300

B.

286

C.

310

D.

240

#### SOLUTION

Solution : B

34+32+30+....+10
This sequence is an AP.
Here, a=34,d=3234=2.
Let the number of terms of the AP be n.
We know that
an=a+(n1)d.

10=34+(n1)(2)

(n1)(2)=24

n1=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.

Find the sum to 100 terms of the series:
1 + 4 + 7 + 5 + 13 + 6 ...

A. 7825
B. 8825
C. 9825
D. 10825

#### 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+(n1)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

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

A. 38
B. 32
C. 6
D. 3

#### SOLUTION

Solution : D

The nth term of an A.P, with first term a and common difference d is given by
Tn=a+(n1)d.
a11=a+10d=35...(1)
a13=a+12d=41...(2)
Solving equations (1) and (2), we get
2d=6.

d=3

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?

A.

1

B.

12

C.

4

D.

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+(n1)d].

Given, S10=4S5.

102[2a+(101)d]=4×52[2a+(51)d]

102[2a+9d]=4×52[2a+4d]
2a+9d=4a+8d
2a=d
ad=12

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+(n1)d to find nth term of AP, we get
a50=a+(501)d106=a+49d(1)
a3=a+(31)d12=a+2d(2)

Subtracting (1) and (2), we get,
47d=94
d=2

Substituting d in (2), we get,
a+2(2)=12
a=124=8

The 29th term is,
a29=a+(291)d=8+28(2)=8+56=64

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 a10 is the 10th term of an AP.
Using formula an=a+(n1)d   to find nth term of arithmetic progression, we get

a17 =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

In the arithmetic progression given below, find the common difference.
1, a, b, c, 9

A.

0

B.

3

C.

1

D.

2

#### SOLUTION

Solution : D

There are five terms in this AP
ana1 + (n - 1)d
ana1=(n1)d
ana1n1=d

Here, a1=1; n=5; an=9
Therefore the common difference,
d=9151=2