# Why is the sum of n terms of an AP equal to a relation that isconstant for all APs

william1941 | Student

The sum of the first n terms of an AP is given by (a1 + an)*(n/2)

To derive this we first see that the nth term of any AP can be written as a1+ (n-1) d.

Now sum of n terms S= a1+ a2 +a3 … an

=> S= a1+ a1 + d +a1+ 2d …a1 + (n-1) d

Or starting with the last term it can be written as

=> S= an + an-d + an-2d +…an- (n-1) d

Now adding the two forms and changing the order of the terms

=> 2S = a1+ an + a1 + d + an - d + a1+ 2d + an - 2d…a1 + (n-1) d + a1 + an - (n-1)d

all terms with d cancel

=> 2S = n (a1 + an)

=> S= (n/2) (a1+ an)

Therefore the sum of the first n terms is S= (n/2) (a1+ an) for all AP

neela | Student

The sum of n terms of an AP  Sn= {2a1+(n-1)d}n/2 is a relation, connecting the  sum  Sn  with a1, the starting term, n the number of terms and d the common difference.

The sum Sn is dependent on n the variable , a1 , and d in the AP like:

Sn = {(2a1+(n-1)d}n/2 = (d/2)n^2 + (a1-d/2)n  is a  second degree  in n.

Here a1 is the first term , ar = a1+(r-1)d th term of the AP

The relation is unique or constant.

Now let us test:

Let  n= 1.

LHS S1 =1. RHS = {2a1 +a1(1-1)d}*1/2 = (2a1+0*d}1/2 =a1.

S2 = {2a1+a1(2-1)d}/2 = a1+a1+d = a1+(a1+d) = a1+a2 holds good.

S3 = (2a1 +a1(3-1)d)3/2 = 3a1+2d = a1+(a1+d)+(a1+2d) = a1+a2+a3 holds good.

S4 = {2a1+a1(4-1)d}4/2 = 4a1+3d = a1+(a1+d)+(a1+2d)+(a1+3d) = a1+a2+a3+a4.

So  Sn = ( d/20n^2+(a1-d/2)n a second degree function (or relation) in n holds true for more than 2 values. So the relation is unique or an identity. Therefore the relation is constant.

krishna-agrawala | Student

In maths as well as in sciences it is possible to express the quantity or values of one variable in terms of function of one or more other terms. This may be called a relationship or a formulas. It is possible to express a variable called a dependent variable in terms of one or more other variables, called independent variables, whenever a fixed and known fixed relationship exists between the dependent and independent variable. Sum of an arithmetic progression (AP) is such an dependent variable, the value of which depends entirely and completely upon three independent variables. These are:

• Value of the first term (a)
• Common difference between terms (d)
• Total number of term (n)

Thus:

Sum of AP = n*a + (n - 1)*n*d/2

Please not that it is possible to have many valid formulas for sum of AP. This is because of the differences in the way the independent variables are defined.