In a A.P , 1st term is a and last term is l . In the A.P term's sumtotal is S. Prove that the common differance is-

The formula for the sum of the terms is:  S  =  n(a + l) / 2

The formula for the n^th (or last) term is:    l  =  a + (n - 1)d

Since you want to have a formula for the common difference that does not include the variable  n,  lets solve the top equation for n and then substitute that value into the second equation.

S  =  n(a + l)/2     --->     2S  =  n(a + l)     --->     2S / (a + l)  =  n

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

Solve the second equation for d:    (l - a) / ( [ 2S / (a + l) ] - 1 )   =   d

Rewrite   [ 2S / (a + l) ] - 1  with the common denominator of (a + l)     (where  1  becomes  (a + l) / (a + l) )

     --->     [ 2S / (a + l) ] - (a + l) / (a + l)

Factor out the denominator of (a + l)

     --->     [ 2s - (a + l) ] / (a + l)

Look back at the left-hand side of the equation for  d  and replace  ( [ 2S / (a + l) ] - 1 )  with   [ 2s - (a + l) ] / (a + l):

     --->     (l - a) / ( [ 2S / (a + l) ] - 1 )   =   (l - a) / { [ 2s - (a + l) ] / (a + l) }

Multiply both the numerator and denominator of this expression by (a + l):

     --->     (l - a)(a + l)  /  { [ 2s - (a + l) ] / (a + l) · (a + l) }      (the (a + l)s on the right will cancel)

     --->     (l - a)(a + l)  /  [ 2s - (a + l) ]

Simplifying and interchanging, this becomes:  ( l² - a² ) / [ 2s - (l + a) ]   which equals  d.

Very nice, geno....!!!!!