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