Test: Arithmetic Progressions

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This mock test of Test: Arithmetic Progressions for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: Arithmetic Progressions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Arithmetic Progressions quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Arithmetic Progressions exercise for a better result in the exam. You can find other Test: Arithmetic Progressions extra questions, long questions & short questions for JEE on EduRev as well by searching above.

QUESTION: 1

If “a” is the first term and ℓ is the last term (nth term), then the sum of all the term of this sequence is given by:

A.
S = n/2 [a + ℓ]
B.
S = n/2 [a + (n - 1)ℓ]
C.
S = [a + (n- 1)ℓ]
D.
S = 2n [a + ℓ]

Solution:

By formula
Sn = n/2(a+l)

QUESTION: 2

Find the sum of first hundred even natural numbers which are divisible by 5.

A.
55005
B.
50500
C.
50000
D.
49500

Solution:

Even natural numbers which are divisible by 5 are 10,20.30...
Difference between the consecutive terms is same
∴ They form an AP whose first term (a) = 10,
common on difference (d) = 10,
and number of terms (n) = 100

∴ The sum of first hundred even natural numbers which are divisible by 5 is 50500.

QUESTION: 3

If the 10 times of the 10th term of an AP is equal to 15 times to the 15th term, then the 25th term is:

A.
25 times of 25th term
B.
zero
C.
1
D.
2

Solution:

10 x 10th term = 15 x 15th term
let a is the first term and d is the common difference .
10(a+9d) = 15(a+14d)
5a + 120d = 0
a + 24d = 0
now , 25th term = a + (25-1)d
= a+24d = 0
hence 25th term = 0

QUESTION: 4

Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....)

A.
64
B.
54
C.
56
D.
81

Solution:

The series is 1², 2², 3², 4², 5², 6², 7², ...
Hence, next term = 8² = 64

QUESTION: 5

The first negative term of the A.P.62,57,52…. is the

A.
10th term
B.
14th term
C.
12th term
D.
18th term

Solution:

a = 62 d = 57 - 62 = -5
tn = a + (n-1)d
= 62 + (n-1)(-5)
= 62 - 5n + 5
= 67 - 5n
From the options, we take '14'
= 67 - 5(14)
= 67 - 70
= -3 (The first negative term will be at the 14th term)

QUESTION: 6

How many terms of the series 24,20,16,…are required so that their sum is 72?

A.
8 or 6
B.
12
C.
9 or 4
D.
11

Solution:

The series is 24,20,16...in AP
so a = 1st term=24 & common difference = 20-24 = -4
sum = 72 is of say n terms
then 72 = n/2[2*24+(n-1)*-4] = n/2[48-4n+4] = n/2[52-4n]
or 144 = 52n-4n²

or 4n²-52n+144 = 0
or n²-13n+36 = 0
or n²-9n-4n+36 = 0
or n(n-9)-4(n-9) = 0
or (n-9)(n-4) = 0
or n = 9 or 4

QUESTION: 7

Roots of quadratic equation x² – 3x = 0 , will be

A.
3
B.
0, –3
C.
0, 3
D.
none of these

Solution:

Given x ² - 3x = 0
Factor x out in the expression on the left.
x (x - 3) = 0
For the product x (x - 3) to be equal to zero we nedd to have
x = 0 or x - 3 = 0
Solve the above simple equations to obtain the solutions.
x = 0
or
x = 3

QUESTION: 8

The number of terms in the sequence -17, -10, -3,…., 144 is:

A.
26
B.
25
C.
27
D.
24

Solution:

QUESTION: 9

Three terms in A.P. are such that their sum is 45. What is the middle term?

A.
25
B.
30
C.
15
D.
10

Solution:

Let the three numbers be a-d, a, a+d
ATQ,
a - d + a + a + 2d = 45
3a = 45
a = 15
Middle term is 15

QUESTION: 10

The terms of an A.P. are doubled, then the resulting sequence is

A.
An A.P. with common difference equal to the common difference of the original A.P
B.
An A.P. with common difference thrice the common difference of the original A.P.
C.
An A.P. with common difference double the common difference of the original A.P.
D.
An A.P. with common difference half the common difference of the original A.P.

Solution:

The general form of an AP is a, a+d, a+2d,.....
where a is the first term and d is the common difference
If we double the terms, the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d nth term= 2a+(n-1)2d
= 2[a+(n-1)d]