\[\large{\color{Blue}{T}\color{Yellow}{U}\color{Magenta}{T}\color{Pink}{O}\color{MidnightBlue}{R} I\color{Blue}{A}\color{Green}{L}!}\]
\[\color{Red}{Arithmetic Progression (AP):}\]
A sequence is in arithmetic progression when its next term to previous term ‘difference’ is always constant. As an example, if I say a frog jumps every 2 seconds starting right now which we take as t=0 sec, the next jump will be on t=2 sec.
When will the next jump be?
Yes, you are correct, it will be at t=4 sec.
Now how did u get 4 ?
Let me tell you, you added the difference d=2 to previous term (2).
So to get next term of AP you always add the common difference to current term, and to get previous term you always subtract the common difference from current term.
A more general formula to get the ‘n’ th term of an AP is:
\[T_n=a_1+(n-1)d\]
Where a_1is the first term and d is the common difference of the AP.
So the general terms of AP are
\[ a_1, a_1+d,a_1+2d,a_1+3d,…..a_1+(n-1)d.\]
When needed to find the sum of ‘n’ terms of an AP, you use:
\[S_n=\frac{n}{2}(2a_1+(n-1)d)\]
But you have a shortcut formula, when you have the 1st and the last term of AP:
\[S_n=\frac{(a_1+a_n)}{2}\] ,where an is the last term.
The sum of infinite Arithmetic Sequence is essentially infinite. But, if a=b=0,the sum is 0.
The product of ‘n’ terms of a finite AP is
\[d^n \frac{\Gamma(a_1/d+n)}{\Gamma(a_1/d)}\]
when a1/d is positive integer.
Note: Ignore this if you are unfamiliar with Gamma Functions.
\[\color{Red}{Geometric Progression (GP):}\]
A sequence is in arithmetic progression when its next term to previous term ‘ratio’ is always constant.
Classical example of Geometric progression is population growth.
If I say population doubles every 15 years, what does it mean?
If the population is 7 billion in 2012, the population in 2027 will be 14 billion.
So what will be the population in 2042?
Yes! you are again correct it will be 28 billion.
How did u get 28?
You multiplied 14 by common ratio 2 to get 28.
So to get next term of GP you always multiply the common ratio to current term, and to get previous term you always divide the current term by common ratio.
A more general formula to get the ‘n’ th term of a GP is:
\[T_n=a_1r^{n-1}\]
Where a1 is the first term and r is the common ratio of the GP.
So the general terms of GP are
\[a_1,a_1r,a_1r^2,…..a_1r^{n-1}.\]
When needed to find the sum of ‘n’ terms of a GP, you use:
\[S_n=a_1\frac{(1-r^n)}{(1-r)}\] when |r|<1
\[S_n=a_1\frac{(r^n-1)}{(r-1)}\] when |r|>1
The sum of infinite geometric sequence with |r|>1 , is essentially infinite.
The sum of infinite geometric sequence with |r|<1 , is
\[S_ \infty=\frac{a}{(1-r)}\].
The product of ‘n’ terms of a finite GP is
\[P=(a_1a_{(n+1)})^{\frac{(n+1)}{2}}\]
\[\color{Red}{Harmonic Progression (HP):}\]
I would call HP as reciprocal of AP, because if we take the reciprocal of the terms in AP, the new terms would be in HP or equivalently, if we take the reciprocal of the terms in HP, the new terms would be in AP.
The general terms of HP are :
\[a_1,\frac{a_1}{(1+d)},\frac{a_1}{(1+2d)}….\frac{a_1}{(1+(n-1)d)} \]
There is no accurate simple general formula for sum to ‘n’ terms of the series.
______________________________________________________________________
Now, lets play with only 3 term sequences.
If a,b,c are in AP, then b is the arithmetic mean(AM) of a and c and is given by
\[b=\frac{(a+c)}{2}\]
This you get from the fact that the difference between consecutive terms are equal,so b-a=c-b or a+c=2b or b=(a+c)/2
If A,B,C are in GP, then B is the geometric mean(GM) of A and C and is given by
\[B=\sqrt{AC}\]
This you get from the fact that the ratio between consecutive terms are equal,so A/B=B/C,so AC=B^2
So B=sqrt(AC).
If X,Y,Z are in HP, then Y is the harmonic mean(HM) between X and Z given by
\[Y=\frac{2XZ}{(X+Z)}\]
Also, this relation is very useful :
The geometric mean of 2 numbers is the geometric mean of arithmetic mean and harmonic mean of those two numbers.
Meaning:
\[GM^2=AM.HM\]
Or here
\[B^2=b.Y\](if a=A=X,c=C=Z)