Arithmetic Progression: Sum of n terms

So, I'm continuing in this post with AP.

    I hope you've gone through our previous post which was focused on arithmetic sequence and Basics,

• If not, click here for reading that post.

 • So, I've revealed the formula of last term of an arithmetic sequence in that post, Now here I'm going to share the formula of finding the sum of n terms of an arithmetic sequence.

Sum of "n" terms of an Arithmetic sequence

  •  Finding Sum of given terms of an AP:-
let there are n terms in given progression and the sum of those terms are defined​ as "S" and formulated as,

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

all other values are known i.e.
"a" as first term and
"d" as common difference.

  • This can formula can also be utilised for evaluating other unknown values if sum is given. That means in this there four (S, a, n, d) variables and any one can be calculated if other three are given.

• Another form of sequence formula

yes, this formula has an another formation too. 

If we have last term of AP i.e. (Tn or l) then formula can be shortened like

S = n/2 [ a+ Tn]

  •  again, this formula can be modified for finding sum of "n" consecutive numbers:

 S= [ n (n+1)] / 2

this formula can be derived as below:

we have 

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

for n consecutive number 

a= 1,  d= 1

so,

S = n/2 [ 2×1 + (n-1)×1]

   = n/2 [ 2  + n-1]

   = n/2 [ n+1 ] 

 hence,

S = [n( n+ 1 )] / 2

you can find an another way of derivation of this sequence formula click here for derivation which is quite easy and it is fun to learn in that method, I personally recommend this post read this at least once.

And also comment about this tutorial how do you feel about this and what else do you want to add here.