CAT Lessons-Progressions
======================

Progressions is a comparatively easy section that comes in CAT.
Once you understand the concepts in progressions, you can answer almost all the questions. Identifying that the questions belong to progressions is the most difficult part. Now we can go to the detailed study of the Progressions.
For a person aiming very high score in QA section,I would advise that he should definitely do the Progressions questions

Progressions can be divided into
------------------------------------------
Arithmetic Progression (AP)
Geometric Progression (GP)
Harmonic Progression. (HP)
Arithmetic Geometric Progression. (AGP)



Arithmetic Progression (AP)
-------------------------------------------

In Arithmetic Progression, the successive terms always has a constant difference between them. The example of an AP is given below.

1,3,5,7,9…….This is an AP series with first term (a or t1)=1 and the constant difference (common difference or cd)=2.
2,4,6,8,10…… This is an AP series with first term (a or t1)=2 and the constant difference (common difference or cd)=2.

The nth term for an AP with first term =‘a’ and cd=’d’ is
tn =a+(n-1)d.
The sum of first n terms of an AP is
Sn=n/2{2a+(n-1)d}


Take the case of AP,2,4,6,8,10……

The 5th term can be found out by
t5=2+ (5-1)2
=2+4*2
=2+8
=10.
As we can see from the series the 5th term is indeed 10.

The sum of first 5 terms can be found out by
S5=5/2{2*2+(5-1)*2)
=5/2{4+8}
=5*6
=30.

Also we can see that sum of first 5 terms are 2+4+6+8+10=30.

Arithmetic Mean (AM)
----------------------------------
AM of an AP with terms a and b is AM = (a+b)/2.
Between two terms of an AP we can put any number of AMs.

Eg.
Suppose we need to put 4 AMs between a and b.,
Now the total number of terms in AP is 6.
Now the common difference in the AP is (b-a)/(4+1)= (b-a)/5..


Properties useful in solving questions in APs
----------------------------------------------------------------------
****
If in an AP, mth term is n and nth term is m, then (m+n)th term is always zero.
Eg. Consider the AP 4,3,2,1,0,-1,-2…..
1st term is 4 and 4th term is 1,therefore (4+1) term ie 5th term ,according to the property should be 0.We can see that the property holds true here.

****
If in an AP, sum of first m terms is equal to sum of first m tems, then sum of first (m+n) term is always zero.
Eg. Consider the AP -2.-1,0,1,2,…..
Sum of first 1 term is -2.
Sum of first 4 terms is -2.
Therefore according to property, sum of first (1+4) ie 5 terms should be 0.
We can verify that the property holds true.

Geometric Progression (GP)
----------------------------------------

In Geometric Progression, the successive terms always has a constant ratio between them. The example of an GP is given below.

1,3,9,27,81…….This is a GP series with first term (a or t1)= 1 and the constant ratio (common ratio or cr)=3.
2,4,8,16,32…… This is a GP series with first term (a or t1)= 2 and the constant ratio (common ratio or r)=2.

The nth term for an GP with first term =‘a’ and common ratio=’r’ is
tn =ar^(n-1)
The sum of first n terms of a GP is
Sn=a{r^(n-1)/(r-1)} r>1
&
Sn = a{r^(n-1)/(1-r)} r<1

Geometric Mean (AM)
------------------------------
GM of a GP with terms a and b is GM = (a*b)^0.5
Between two terms of an GP we can put any number of GMs.

Eg.
Suppose we need to put 4 =GMs between a and b.
Now the total number of terms in =GP is 6.
Now the common ratio of the GP is (b/a)^(4+1)= (b/a)^0.5

But in Competitive exams like CAT,GMAT etc, in GP, sum to infinity is more significant than sum of n terms.
Sum to infinity of GP is infinity if r>1,so the cases where, r<1 only we will get a definite sum.
Sum to infinity of a GP with first term,a and common ratio,r is
S∞=a/(1-r).

Eg. Find the sum to infinity of the series 1,1/2,1/4,1/8,1/16…….
a=1,r=1/2
S∞=1/(1-1/2)=2.

Harmonic Progression
-----------------------------------

If a,b,c are in Harmonic Progression(HP),then 1/a,1/b,1/c are in AP.
This is the standard definition of an HP.


Also we have the general rules
1. AM>=GM>=HM
2. GM^2=AM*HM



Strategies for solving problems in Progressions.
----------------------------------------------------------------------

**
In all questions that has Tn and Sn variable, n, always put values for n and check for the answer option.

**
Always try to make up a progression satisfying the conditions given in the questions and solve that progression only instead of doing with unknown numbers.

**
Suppose we need to take one AP,one GP and one HP having 3 terms with same first and last terms,then we an take the following progressions,

AP: 1, 2/3,1/3 cd=1/3
GP : 1,1/(3^.5),1/3 cr=1/(3^.5)
HP: 1,1/2,1/3 HM=1/2.

**
Find the 625th term of the series 1,2,2,3,3,3,4,4,4,4,5………

Analysing the series, we see

Last 2 is the 3rd term ie 2*3/2
Last 3 is the 6th term ie 3*4/2
So we got the pattern,Last n will be the n(n+1)/2 th term
Also the previous n terms will be same,n only.

So if we consider,the case of 625th term, consider 35*36/2 gives 630.
Means 630th term will be 35,also previous 35 terms will be same 35 only.
So our answer is 625th term is 35.