Quantitative Aptidue,Progression questions
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Progression

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Progression  Aptitude basics, practice questions, answers and explanations 
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Important Equations and Formula

Sum of first n natural numbers= n(n+1)/2
Sum of the squares of first n natural numbers= (n(n+1)(2n+1))/6
Sum of the cubes of first n natural numbers= [n (n+1)/2]2
Sum of first n natural odd numbers= n2
Average = Sum of items/ Number of items

Arithmetic Progression (AP): An AP is of the form a, a+d, a+2d, a+3d,..... where a is called the 'first term' and d is called the 'common difference'.
nth term of an AP; tn=a+(n-1)d
Sum of the first n terms of an AP; Sn= n/2 [2a+(n-1)d] or Sn= n/2 (first term+last term)

Geometrical Progression (GP):
A GP is of the form a, ar, ar2, ar3......... where a is called the 'first term' and r is called the 'common ratio'.
nth term of a GP; tn= arn-1
Sum of the first n terms in a GP; Sn= a(1-rn)/1-r
Sum of infinite series of progression; S= a/(1-r)
Geometric mean of two number a and b is given as GM= sqrt(ab)

Harmonic Progression (HP)

If a1,a2,a3,...................an are in AP, then 1/a1, 1/a2, 1/a3, ........1/an, are in HP
nth term of this HP, tn =1/(1/a1+(n-1)(a1-a2/a1a2) ) nth term of this HP from the end, tn=1/ (1/a1-(n-1)(a1-a2/a1a2))

If a and b are two non-zero numbers and H is harmonic mean of a and b then a, H, b from HP and then H=2ab/(a+b)

Arithmetico-Geometric series

A series having terms a, (a+d)r, (a+2d)r2,...... etc is an Arithmetico-Geometric series where a is the first term, d is the common difference of the Arithmetic part of the series and r is the common ratio of the Geometric part of the series.

The nth term tn= [a+(n-1)d]rn-1
The sum of the series to n terms is
Sn= a/1-r+ dr (1-rn-1)/ (1-r)2 - [a+(n-1)d]rn/ 1-r
The sum to infinity, S= a/ 1-r + dr (1-rn-1)/(1-r)2 ; r<1

Exponential Series

ex = 1+x/1!+x2/2!+x3/3!+..........   aptitudeprogression(e is an irrational number)
coefficient of xn= 1/n!; Tn+1=xx/n!
e-x = 1-x/1!+x2/2!- x3/3!+..........
aptitudeprogression

Logarithmic Series

loge (1+x)= x-x2/2+x3/3+x4/4+........  aptitudeprogression(-1<x aptitudeprogression1)

loge (1-x)= -x-x2/2-x3/3- x4/4 -........  aptitudeprogression(-1x<aptitudeprogression 1)

loge (1+x)/(1-x)= 2-(x+x3/3+x5/5+.........) aptitudeprogression(-1<x aptitudeprogression1)