Can someone help to prove the sum sn of the nterm of the A.P?

Answer 1

Yes. You're asking for the sum Sn of n terms of an arithmertic progression (or series, as we more generally call it). Here is how to find/prove what that is:

Let there be ' n ' terms, let the first term be ' a ', and the common difference ' d '; then the sum of n terms, Sn is:

Sn = a + [a + d] + ... + ... +... + [a + (n - 1) d].

Rewrite this with the terms on the RHS all in the opposite order:

Sn = [a + (n - 1) d] + ... + ... + ... + a.

Now add these two equations. (Writing this by hand, you can arrange for things to line up nicely, vertically, but they do not do that so nicely here because of font spacing limitations.) Each VERTICAL PAIR of terms on the RHS sum to the same amount, [2 a + (n - 1) d]. There are ' n ' of these vertical pairs. Thus:

2 Sn = n [2 a + (n - 1) d], therefore Sn = (1/2) n [2 a + (n - 1) d]. QED

Live long and prosper.

Answer 2

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

Answer 3

You've got to be more specific

Answer 4

t(n) = a + (n-1)d
s(n) = (n/2)(a + t(n))

Th