TOPICS TO BE COVERED :
–>nth term of an AP
–>sum of AP upto n terms
–>nth term of a GP
–>sum of GP upto n terms
Let’s start with a game of sequences, basically you have to tell the next term of these sequences :
Of course we can easily guess the next terms as
We just observe that every next term is obtained by adding a constant number to the previous term.For example, adding 2 to 2 gives us 4,again adding 2 to 4 gives us 6,so the next term must be obtained by adding 2 to 6 which is ,of course 8.
Did you know the sequences given above are nothing but arithmetic progressions ?
Arithmetic is a branch of mathematics relating to numbers and counting.
Progression means gradually increasing.
So,arithmetic progression (AP)means a sequence which can be increased by adding a constant to the previous term.
Now,the question arises that can we find any term of an AP?
Of course we can !!
Let’s observe the pattern of an AP given below:
The first term is 3.
The second term is 5,which is 3+2.
The third term is 7,which is 5+2 or 3+2+2 or 3+2(2).
The fourth term is 9,which is 7+2 or 3+2+2+2 or 3+3(2).
Let us denote the first term as ‘a’.
Now,every AP has a constant which can be added to previous term to get the next term,we call it as common difference and denote it by ‘d’.
WHY DO WE CALL IT AS COMMON DIFFERENCE ?
Because it is the difference between the two consecutive terms of the AP.
For example, here 7-5=2 and 9-7=2 ,so d=2.
Now,if we talk in general, we observe that the second term of an AP is given by a+d,
The third term is given by a+2d,
Similarly,the nth term is given by a+(n-1)d.
We denote the nth term as an.
So,we have found a way to get any term of an AP by just knowing it’s first term and the common difference.
WHAT IF WE WANT TO FIND THE SUM OF AP UPTO N TERMS ?
Let’s again observe an AP:
Sum upto first term : 3,which is a.
Sum upto second term : 8,which is 3+5,which is a+(a+d) or 2a+d
Sum of three terms :15,which is 3+5+7,which is a+(a+d)+(a+2d) or 3a+3d
Sum of four terms : 24,which is 3+5+7+9,which is a+(a+d)+(a+2d)+(a+3d )or 4a+6d.
If we guess the sum of five terms,it will be 5a+10d.
If we just observe the coefficients of ‘d’,we will notice that it follows a pattern as : 1,3,6,10……..
As we know about the combination formula,we see that
4C2=6, and so on…..
So,let’s come to conclusion !!
The sum of n terms of an AP is given by
na + nC2 d.
Please note that we can find out the sum of n terms of AP without using the combination formula also,
We have 1,3,6,10……so on.
Now, 1= 2(2-1)/2
So,for n, it is n(n-1)/2.
So,the sum of n terms can be given by
an + n(n-1)/2.
If we multiply and divide by 2, and take n common,we have
We denote the sum of n terms as Sn.
WHAT IF WE WANT TO FIND THE LAST TERM OF A FINITE AP?
If we know the number of terms,then we can find the last term easily by using the simple formula.
Suppose an AP have n no. of terms,then the last term,denoted by ‘l’ is given by a+(n-1)d.
IF WE WANT TO FIND THE SUM OF AP UPTO THE LAST TERM,
We can simply apply the formula and get the answer.
Now,if we are given the first and last term,we still can find the sum of AP,
Sn= n/2[2a + (n-1)d]
Suppose we have two numbers : 3,5
If we want to insert a number between these two so that the three numbers form an AP,we can do so by inserting 4 .
HOW DID YOU KNOW THAT THE NUMBER IS 4?
It’s simple, if we assume the number to be d, then d-3 = 5-d,
So, d+d = 5+3
: 2d = 8,
: d = 4.
Did you know that this number is called the
arithmetic mean of these two numbers ?
So, if we have any two numbers a,b then the arithmetic mean,AM = (a+b)/2.
Now,try to find the next term of these sequences:
The answers are of course:
32 and 81,
HOW DID YOU ANSWER?
We just observe that 2X2=4,4X2=8,8X2=16,So the next term must be obtained by multiplying 16 by 2 which is 32 and it’s true.
DID YOU KNOW THAT THESE SEQUENCES ARE CALLED AS GEOMETRIC PROGRESSIONS!!
Yes!! Each term of a GP is obtained by multiplying the previous term by a fixed non- zero number,which is called as the common ratio.
WHY IT IS CALLED AS THE COMMON RATIO?
Because by taking the ratio of any two consecutive terms,we get it.
For example,look at the following GP:
If we divide 16 by 4,we get 4,and if we divide 64 by 16,we get again 4.
So,here ,the common ratio,’r’ is 4.
WHAT IF WE WANT TO FIND ANY TERM OF A GP?
Let’s observe the following GP:
Let us denote the first term as ‘a’.
Second term = 2, which is 1X2 or a(r)
Third term = 4,which is 1X2X2 or a(r)2
Fourth term = 8,which is 1X2X2X2 or a(r)3.
nth term= a(r)n-1.
We denote it by an=a(r)n-1.
WHAT IF WE WANT TO FIND THE SUM OF GP UPTO N TERMS?
Let’s observe the GP given below:
Sum upto first term: 2,or a.
Sum upto second term:8,which is 2+6,or a+ar which is a(1+r).
Sum upto third term:26,which is 2+6+18,or a+ar+a(r)2,which is a(1+r+r2).
Sum upto fourth term: 80,which is 2+6+18+54, or a+ar+ar2+ar3
Or a(1+r+r2+r3) .
WHAT WE OBSERVED ?
We see that there is no pattern like AP here to find the sum upto n terms.
SO NOW ?
Don’t worry !!
We have another method to find the sum. This method is often prefered for discovering new formulas if a fixed pattern is not visible.
Now, we have
Sn = a + ar + ar2 +…….+ arn-1. Equation 1)
If r=1, Sn = a+a+a…….upto n terms = na.
Let’s multiply equation 1) by r,
So, rSn = ar + ar2 + ar3 + ……. arn. Equation 2)
EQUATION 2) FROM EQUATION 1),
Sn – rSn = a- arn.
Sn(1-r) = a(1-rn)
So, Sn = a(1-rn)/(1-r).
Please note that we can also write above as :
Sn = a(rn-1)/(r-1).
This is because a/b is same as –a/-b and x-y = -(y-x).