# Free Arithmetic Progressions 02 Practice Test - 10th Grade

The sequence of numbers 5, - 2, -9, -16, ... is

A. an AP with common difference -7.
B. an AP with common difference 7.
C. an AP with common difference 3.
D. not an AP.

#### SOLUTION

Solution : A

For a sequence to be an AP, the same common difference must exist between any two consecutive terms of AP.

Here,
a1=5,a2=2 and a3=9

a2a1=25=7
a3a2=9(2)=7

The difference between any two consecutive terms is -7.

Hence, the list of numbers are in AP with common difference -7.

If 2x, x + 10, 3x + 2 are in AP. Find the value of x.

__

#### SOLUTION

Solution :

Since, 2x, x + 10, 3x + 2 are in AP.

2 (x + 10) = 2x + (3x + 2)

2x + 20 = 5x + 2

3x = 18

x = 6.

How many terms are there in the following AP?
7, 10, 13, …, 43

A. 12
B. 14
C. 16
D. 13

#### SOLUTION

Solution : D

Here
First term, a = 7
Common difference, d = 10 - 7 = 3

Using the formula for nth term
an=a+(n1)d
43=7+(n1)3
36=(n1)3
12=n1
n=13
There are total 13 terms in the given AP.

Find the number of terms in each of the following APs:

(i) 7,13,19....,205

(ii) 18,312,13,...47

A.

34, 27

B.

26, 35

C.

27, 34

D.

35, 26

#### SOLUTION

Solution : A

(i)  7,13,19....,205

First term, a=7

Common difference, d=137=6

an=205

Using formula an=a+(n1)d to find nth term of arithmetic progression, we get

205=7+(n1)6
205=6n+1

204=6n

n=2046=34

Therefore, there are 34 terms in the given arithmetic progression.

(ii) 18,312,13,...47

First term, a=18

Common difference, d=31218=52

an = − 47

Using formula an=a+(n1)d to find nth term of arithmetic progression, we get

47=18+(n1)(52)

94=365n+5

5n=135
n=1355=27

Therefore, there are 27 terms in the given arithmetic progression.

Which term of the AP : 3, 8, 13, 18, . . . , is 78?

___th

#### SOLUTION

Solution :

Here, First term, a=3

Common difference, d=83=5
an=78
Using formula an=a+(n1)d to find nth term of arithmetic progression, we get
an=3+(n1)5  {we want to find value of n here.}
78=3+(n1)5
75=5n5
80=5n
n=805=16
It means 16th term of the given AP is equal to 78.

A spiral is made up of successive semicircles, with centers alternatively at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in the figure.What is the total length of such a spiral made up of thirteen consecutive semicircles?

__ #### SOLUTION

Solution :

Length of semi-circle =circumference of circle2 = 2πr2 =  πr .

Length of semi-circle of radii 0.5 cm = π×0.5 cm

Length of semi-circle of radii 1.0 cm = π×1.0 cm

Length of semi-circle of radii 1.5 cm = π×1.5 cm

Length of semi-circle of radii 2.0 cm = π×2.0 cm

.....and so on

π(0.5),π(1.0),π(1.5).....   13 term (There are total of thirteen semi-circles}.
For total length of the spiral, we need to find sum of the sequence   13 terms
Total length of spiral = 0.5π+π+1.5π+2π........upto 13 terms

Total length of spiral = π(0.5+1.0+1.5+.......) upto 13 terms

Sequence 0.5, 1.0, 1.5 ....13 terms is an arithmetic progression.

Let's find the sum of this sequence.

a=0.5;d=0.5

S=n2(2a+(n1)d)

S=132(2×0.5+(131)(0.5))

S=132(1+6)

S=912=45.5

So

Total length of spiral = π(0.5+1.0+1.5+.......)

Total length of spiral = π(45.5)=143 cm

Find the sum to n terms of the AP:
5, 2, -1, -4, -7, ...

A.

n2(133n)

B.

n(23n)

C.

n2

D.

n(n2)

#### SOLUTION

Solution : A

The given sequence is  5, 2, -1, -4, -7, ...., an AP with first term a=5 and common difference d=25=3.

The sum to n terms of an AP of first term a and common difference d is
Sn=n2[2a+(n1)d].Sn=n2[10+(n1)(3)]
=n2[103n+3]
=n2[133n]

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

A.

568

B.

624

C.

564

D.

672

#### SOLUTION

Solution : D

As an=3+2n,

so,

a1=3+2×1=5

a2=3+2×2=7

a3=3+2×3=9

List of numbers becomes 5, 7, 9, 11 . . .

Here, 75=97=119=2 and so on.

So, it forms an AP with common difference d=2.

To find S24, we have n=24, a=5, d=2.

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

=12(10+46)=672

If the sum of p terms of an AP is q and the sum of q term is p, then the sum of p + q terms will be:

A. 0
B. p - q
C. p + q
D. -(p + q)

#### SOLUTION

Solution : D

let first term be a and common difference be d
so pthterm=a+(p1)d
and sum of first p terms=p2(2a+(p1)d)=q
hence(2a+(p1)d)=2qp....(1)and qthterm=a+(q1)d
and sum of first q terms=q2(2a+(q1)d)=phence(2a+(q1)d)=2pq....(2)subtract (1) from (2)(pq)d=2(qppq)=2(q2p2)qso,d=2(p+q)pq....(3)(p+q)thterm=(a+(p+q1)d)and sum of its first(p+q)term=(p+q)2(2a+(p+q1)d)=(p+q)2(2a+(p1)d+qd)=(2qp+qd)(p+q)2...from(1)=(2qp22qp)p+q2...from(3)=(p+q)

What is the arithmetic mean of the numbers a and b?

A.

a+bab

B.

a+b2

C.

a+b

D.

ab2

#### SOLUTION

Solution : B

The arithmetic mean of two numbers is the simple average of the two numbers.

Thus,

AM=Sum of the numbersNumber of numbers

=a+b2