Arithmetic progression I Arithmetic Progression for class 10th
                     What is arithmetic progression?
An arithmetic progression can be understood as a sequence that has a  
starting number and a series advancing with a definite interval. The simplest  
and easiest example being the number series 1,2,3,4,5,6,7. Here, the first  
number is 1 and the definite interval is 1. If we keep on adding 1 to the last  
number, the sequence will keep on progressing and yes, it becomes an  
arithmetic progression!
Most used terms in arithmetic progression lesson
Now, let us transform this definition into a mathematical formula to sound  
more studious. Let us name all the terms first. (Most used terms in arithmetic  
progression lesson)
When the sign of = comes in picture, an expression becomes an equation. So,  
the expression 2+4x=6 becomes an algebraic equation.
First term be a1,
Difference be d; and 
 Last term be an
Now, add ‘d’ to a1. What do you get? (a1+d). It means you got two numbers of  
the sequence. The two numbers are a1 and (a1+d).
a1 = a1
a2 = (a1+d)
a3 = (a1+d) + d 
 a3 = a1 + 2d
Similarly a4 
can be written 
as a4 = a1 + 
3d;
In the same manner, if you need to find any number of the given arithmetic  
progression, you can simply apply this trick, and lo, the mathematicians call it  
a formula. Did you guys realize? You have just made up formula. So, here  
goes the formula
To find the nth term of any progression?
an = a1 + (n-1)d
Sounds simple right? Well, yes it is. Let’s check it out with the help of one  
example. My birthdate is 6 so let the first term be 6. And my birth month is 3,  
so let the difference be 3. We have got a1=6 and d=3. Let us put it into the  
formula and find out the 11th term of the series.
an = a1 + (n-1)d 
 a11=6+(10*3)  
a11=36
Bingo! As easy 
as that.
Now that you have learned how to calculate any number in a given arithmetic  
progression, let’s proceed onto the next formula.
Sometimes, finding the nth term is not enough and you need to find the sum of  
a definite arithmetic progression. How to go about with that?
The sum of arithmetic progression
The sum of the nth term of any given arithmetic progression can be calculated  
by the sum of the first term and the last term divided by half and multiplied by  
the number of terms in the series. To make it easier, let’s put it into a formula.
Sn=n/2(a1+an)
But this sounds quite simple. When we delve deeper into the subject, you may  
realize that the nth term is not necessarily given every time. In that case, you  
need to depend upon another formula, that is:
Sn=n/2{2a1+ (n+1)d}
If you are wondering how we came up with this formula, let me explain to you.  
We did nothing but substituted the value of an from the previous formula. Let’s  
check out.
Sn=n/2(a1+an)
But, we know an = a1 + (n-1)d
Let us substitute this value in the Sn formula.
Sn=n/2(a1+an)  
Sn=n/2(a1+a1 + (n-1)d) 
 Sn=n/2{2a1 + (n-1)d}
How cool and easy! Let us take a few examples to understand it better. So,  
let’s begin with my favorite numbers. Consider the first sequence where a1=6  
and d=3, and suppose the last value a be 36. Now, putting these in the first  
formula of Sn, we will get;
Sn=n/2(a1+an)
Sn=11/2(6+36) 
 Sn=231
Bravo! Now, let’s proceed with another example where we don’t know what is  
the last number of the sequence. Suppose the value of a1=3 and d=6, we  
need to find the sum of the number till the 20th number of the series. Let’s go!
Sn=n/2{2a1 + (n-1)d}
Sn=20/2{2*3 + (20-1)*6}v Sn=10{6+114}  
Sn=1200
Did you know? After becoming an expert, you can also calculate the amount  
collected in your piggy bank (of course, if you add money at proper intervals  
and don’t cheat). Suppose you initially put 500 in your piggy bank and  
hereafter keep on adding 100 every month, how much money will you have at  
the end of the year? Let’s calculate.
a1=500 
 d=100  
n=12
an = a1 
+ (n-
1)d
an = 
500 + 
(12-
1)100
an = 
1600
So, you will be getting 1600 by the end of the year. Good savings, indeed! In a  
similar fashion, you can calculate the fare of a taxi when you travel. They have  
a fixed initial rate and the rate increases according to the distance you travel.  
So, ask your cab driver the next time you travel and calculate your own fare.
Amazing! Isn’t it? It seems you are becoming an expert in solving arithmetic  
progressions!
To find the sum of an arithmetic progression but 
 from the last term.
Let’s go further and cover up the one last formula as well. It is not any  
different, just a twisted version of the same thing. Suppose you need to find  
the sum of an arithmetic progression but from the last term. It means, you  
need to find the sum of the arithmetic progression but in reverse order. Let's  
try to put it in our formula for finding the sum.
Sn=n/2{2a1 + (n-1)d}
But as we have to find it out from the last term, let’s call it al. So, let’s replace 
 a1 with al and as we have to go in reverse order, the difference d will become 
 (-d).
By replacing all the values, you will 
get;  Sn=n/2{2a1 + (n-1)d}
Sn=n/2{2al + (n-1)(-
d)}  As simple as that.
Just in case you need to understand it better, let’s try to find the sum of the  
first 100 even numbers from the last number. So here, al =100, d=(-2) but we  
don’t know how many even numbers are there from 1 to 100. So, first, we  
need to find the value of n. Let’s start.
Formulae known to us: an = a1 + (n-1)d
For this problem, we have the following values with us  
an=100
a1=2
d=2
Putting these in the equation, we’ll find the value of n.  
an = a1 + (n-1)d
100=2+{(n-1)*2}
98=2n-2
96=2n
48=n
Hence, there are 48 even numbers from 1 to 100. Now, let’s find the sum from  
the last number.
al=100
n=48  
d=(-2)
Sn=n/
2{2al 
+ (n-
1)(-d)}
Sn=n/
2{2al 
+ (n-
1)(-d)}
Sn=48/2{2*100+ (48-1)(-2)} 
 Sn=2,544
See! We have understood the fundamentals of arithmetic progressions like a  
cakewalk. So bring on your witty caps and solve as many problems you come  
across while reading this topic and don’t shy away from this fun-filled, easy,  
and understandable topic. Happy solving!