Progression - Online Quiz


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Following quiz provides Multiple Choice Questions (MCQs) related to Progression. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Questions and Answers

Q 1 - 105 th term of the A.P. 4, 9/2, 5, 11/2, 6 ....is

A - 56

B - 111/2

C - 119/2

D - 55

Answer : A

Explanation

 Here a = 4, d = (9/2-4) = 1/2
T₁0₅ = a+(105-1)*d=4+104*1/2=4+52=56.

Q 2 - In the event that the fourth term of a number juggling movement is 14 and its twelfth term is 70, then its first term is:

A - - 10

B - -7

C - 7

D - 10

Answer : B

Explanation

  Let the first term of the A.P. be a and common difference be d. then,
a+ 3d = 14 ...(i)    a+11d = 70 ...(ii)
On subtracting (i) from (ii), we get 8d =56    d = 7
Putting d = 7 in (i), we get a+3*7=14 ⇒ a= (14-21) = -7   ∴ First term = -7

Q 3 - What number of products of arrive between the whole numbers 15 and 105, both comprehensive?

A - 30

B - 31

C - 32

D - 33

Answer : B

Explanation

Requisite no. are 15, 18, 21, 24, 105.
Here a = 15 and d = (18-15) = 3
Let their number be n. Then,
a + (n-1) d =105 ⇒ 15+ (n-1)*3 = 105
⇒ a+ (n-1) d = 105 ⇒ 15+ (n-1)*3 = 105
⇒ (n-1)*3 = 90   ⇒ n-1 =30 ⇒ n = 31

Q 4 - What is the following number in the math movement 2, 5, 8....?

A - 7

B - 9

C - 10

D - 11

Answer : D

Explanation

This is an A.P. in which a = 2 and d = (5-2) =3.
∴ Next number= (8+3) =11.

Q 5 - What number of odd number pages arrives in a book of 1089 pages?

A - 542

B - 544

C - 545

D - 546

Answer : C

Explanation

The pages are 1, 3, 5, 7...,1089.
This is an A.P. in which a= 1, d = (3-1) = 2 and L = 1089.
Let the number of these term be n. then,
(a+ (n-1) d = 1089 ⇒ 1 + (n - 1)* 2 = 1089 ⇒ (n-1 ) * 2 = 1088
n-1 =544⇒ n = 545.
∴ Required number of pages=545.

Q 6 - What number of term of the series 3,9,27 ... will mean 363?

A - 5

B - 6

C - 7

D - 8

Answer : A

Explanation

Given series is a G.P. in which a =3, r= 3 and Sn =363
∴ Sn = a (rⁿ-1)/(r-1) ⇒ 3*(3ⁿ-1)/ (3-1) = 363 ⇒ 3ⁿ-1 = 363*2/3 = 242
3ⁿ= 243 = 3⁵⇒ n =5

Q 7 - The second term of a geometrical progression is 2/3 and its fifth term is 16/81. Its seventh term is:

A - 15/524

B - 1/32

C - 32/729

D - 64/729

Answer : D

Explanation

Let the G.P. be a, ar, ar2...
Then T2 =2/3 and T₅ = 16/81 ⇒ ar = 2/3 ...(i) And ar⁴= 16/81 ...(ii)
∴ar⁴/ar =16/81*3/2 = 8/27 ⇒ r3 = 8/27 ⇒ r= 2/3
Putting r =2/3 in (i), we get =a= 1.
T7 = ar⁶ = 1* (2/3)⁶ = 2⁶/3⁶ = 64/729

Q 8 - A clock hums 1 time at 1o, clock, 2 times at 2o, clock, 3 times at 3o, clock etc. What will be the aggregate quantities of hums in a day?

A - 100

B - 150

C - 156

D - None of these

Answer : C

Explanation

Total number of buzzes =2 (1+2+3+..........+12).
This is an A.P. in which a=1, d=1, n = 12 and L = 12.
(1+2+3+ .........+12) = 12/2* (1+12) = 78.
∴ Total number of buzzes = (2 * 78) = 156.

Q 9 - (13 +23 + 33+...... + 153) =?

A - 900

B - 11025

C - 13400

D - 14400

Answer : D

Explanation

We know that (13+23+33+??+n3) = {n (n-1)/2)2
∴ (13+23+33+?..+153) = (15*16/2)2 = (120)2 = 14400.3

aptitude_progression.htm
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