**TOPICS TO BE COVERED :**

–>nth term of an AP

–>sum of AP upto n terms

–>arithmetic mean

–>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 :

1,2,3,4,___

2,4,6,___

3,6,9,___

Of course we can easily guess the next terms as

5

8

12

But

**HOW ?**

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 !!

**HOW?**

Let’s observe the pattern of an AP given below:

3,5,7,9……….

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 a_{n}.

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:

3,5,7,9……..

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

^{2}C_{2}=1,

^{3}C_{2}=3,

^{4}C_{2}=6,
and so on…..

So,let’s come to conclusion !!

The sum of n terms of an AP is given by

**na + ^{n}C_{2}
d.**

Please note that we can find out the sum of n terms of AP without using the combination formula also,

### HOW?

We have 1,3,6,10……so on.

Now, 1= 2(2-1)/2

3=3(3-1)/2

6=4(4-1)/2

.

.

So on.

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

**n/2[2a+(n-1)d].**

We denote the sum of n terms as S_{n.}

**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.

Now,

**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,

S_{n}= n/2[2a + (n-1)d]

OR n/2[a+a+(n-1)d]

Now, a+(n-1)d=l

So,

**S _{n}=n/2[a+l]**

**ARITHMETIC MEAN**

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:

2,4,8,16,__

3,9,27,___

YES!

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:

4,16,64…….

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:

1,2,4,8,…….

Let us denote the first term as ‘a’.

So, a=1.

Here, r=2.

Now,

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}.

SO,

nth term= a(r)^{n-1}.

We denote it by** a _{n}=a(r)^{n-1}.**

**WHAT IF WE WANT TO FIND
THE SUM OF GP UPTO N TERMS?**

Let’s observe the GP given below:

2,6,18,54………

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+r^{2}).

Sum upto fourth term: 80,which is 2+6+18+54, or a+ar+ar^{2}+ar^{3}

Or a(1+r+r^{2}+r^{3})
.

**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

S_{n }= a + ar
+ ar^{2} +…….+ ar^{n-1}. Equation 1)

If r=1, S_{n}
= a+a+a…….upto n terms = na.

Let’s multiply equation 1) by r,

So, rS_{n} = ar + ar^{2} + ar^{3} +
……. ar^{n}.
Equation 2)

**SUBTRACTING **

**EQUATION 2) FROM
EQUATION 1),**

S_{n } – rS_{n}
= a- ar^{n}.

S_{n}(1-r) = a(1-r^{n})

So, **S _{n} = a(1-r^{n})/(1-r).**

Please note that we can also write above as :

**S _{n} = a(r^{n}-1)/(r-1).**

This is because a/b is same as –a/-b and x-y = -(y-x).