Quantitative Aptitude - Arithmetic Ability Questions

Arithmetic Series ::

An Arithmetic Series is a series of numbers in which each term increases by a constant amount.

How to find the sum of the Arithmetic Sequence or Series for the given Series ::

When the series contains a large amount of numbers, its impractical to add manually. You can quickly find the sum of any arithmetic sequence by multiplying the average of the first and last term by the number of terms in the sequence.

That is given by Sn = n(a1 + an)2 Where n = number of terms, a1 = first term, an = last term

Here Last term is given by an = a1 + n-1d where d = common difference

Now given Arithmetic Series is 

2, 5, 8, 11,...

Here a1 = 2,  d = 3, n = 36 

Now, $$\begin{gathered} an= a1 + n - 1d \\ a36= 2 + 36 - 13 = 105 + 2 = 107 \end{gathered}$$

Now, Sum to 36 terms is given by

S36 = 36(2 + 107)2 = 36 x 1092 = 39242 = 1962

Therefore, Sum to 36 terms of the series 2, 5, 8, 11,... is 1962.

Given 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28

It is in the form of

 a3+b3+c3-3abca2+b2+c2-ab-bc-ca= a + b + c

Here a = 37, b = 35 & c = 28 

=> a + b + c = 37 + 35 + 28 = 100

Therefore,   = 100

Subtract 1, 3, 5, 7, 9, 11 from successive numbers.

So, 34 is wrong.

All except Parrot are birds of prey.

The numbers are 7 x 8, 8 x 9, 9 x 10, 10 x 11, 11 x 12, 12 x 13.

So, 150 is wrong.

Let A, B, C be the respective events of solving the problem and A , B, C be the respective events of not solving the problem. Then A, B, C are independent event

∴A, B, C are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

 PA=12, PB=23, PC= 34

∴ P( none  solves the problem) = P(not A) and (not B) and (not C)  

                  = PA∩B∩C 

                  = PAPBPC         ∵ A, B, C are Independent                       

                  =  12×23×34  

                  = 14  

Hence, P(the problem will be solved) = 1 - P(none solves the problem) 

                = 1-14= 3/4