# Free Arithmetic Progressions 02 Practice Test - 10th Grade

### Question 1

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

#### 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

a2−a1=−2−5=−7

a3−a2=−9−(−2)=−7

The difference between any two consecutive terms is-7.

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

### Question 2

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.

### Question 3

How many terms are there in the following AP?

7, 10, 13, …, 43

#### SOLUTION

Solution :D

Here

First term, a = 7

Common difference, d = 10 - 7 = 3

Using the formula for nth term

an=a+(n−1)d

43=7+(n−1)3

36=(n−1)3

12=n−1

n=13

There are total 13 terms in the given AP.

### Question 4

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

(i) 7,13,19....,205

(ii) 18,312,13,...−47

34, 27

26, 35

27, 34

35, 26

#### SOLUTION

Solution :A

(i) 7,13,19....,205

First term, a=7

Common difference, d=13−7=6

an=205

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

205=7+(n−1)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=312−18=−52

an = − 47

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

−47=18+(n−1)(−52)

⇒−94=36−5n+5

⇒5n=135

⇒n=1355=27Therefore, there are 27 terms in the given arithmetic progression.

### Question 5

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

#### SOLUTION

Solution :Here, First term, a=3

Common difference, d=8−3=5

an=78

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

an=3+(n−1)5 {we want to find value of n here.}

⇒78=3+(n−1)5

⇒75=5n−5

⇒80=5n

⇒n=805=16

It means 16th term of the given AP is equal to 78.

### Question 6

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+(n−1)d)

⇒S=132(2×0.5+(13−1)(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

### Question 7

Find the sum to n terms of the AP:

5, 2, -1, -4, -7, ...

n2(13−3n)

n(2−3n)

n−2

n(n−2)

#### SOLUTION

Solution :A

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

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

Sn=n2[2a+(n−1)d].⇒Sn=n2[10+(n−1)(−3)]

=n2[10−3n+3]

=n2[13−3n]

### Question 8

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

568

624

564

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, 7–5=9–7=11–9=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)+(24−1)2)

=12(10+46)=672

### Question 9

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:

#### SOLUTION

Solution :D

let first term be a and common difference be d

so pthterm=a+(p−1)d

and sum of first p terms=p2(2a+(p−1)d)=q

hence(2a+(p−1)d)=2qp....(1)and qthterm=a+(q−1)d

and sum of first q terms=q2(2a+(q−1)d)=phence(2a+(q−1)d)=2pq....(2)subtract (1) from (2)(p−q)d=2(qp−pq)=2(q2−p2)qso,d=−2(p+q)pq....(3)(p+q)thterm=(a+(p+q−1)d)and sum of its first(p+q)term=(p+q)2(2a+(p+q−1)d)=(p+q)2(2a+(p−1)d+qd)=(2qp+qd)(p+q)2...from(1)=(2qp−2−2qp)p+q2...from(3)=−(p+q)

### Question 10

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

a+ba−b

a+b2

a+b

a−b2

#### 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