given-the-AP-a-a-d-a-2d-a-3d-show-that-S-n-n-2-2a-n-1-d-

Question Number 45410 by Rio Michael last updated on 12/Oct/18

g i v e n t h e A P

a, a + d, a + 2 d, a + 3 d, \dots

s h o w t h a t

S_{n} = \frac{n}{2} {(2 a + (n - 1) d}

Commented by maxmathsup by imad last updated on 13/Oct/18

w e h a v e u_{n} = a + n d l e t S_{n} = u_{0} + u_{1} + \dots . + u_{n - 1} \Rightarrow

S_{n} = \frac{n}{2} {u_{0} + u_{n - 1}} = \frac{n}{2} {a + a + (n - 1) d} = \frac{n}{2} {2 a + (n - 1) d} .

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

S_{n} = a + (a + d) + (a + 2 d) + . . + {a + (n - 1) d}

S_{n} = {a + (n - 1) d} + {a + (n - 2) d} + . . + a

a d d t h e m

2 S_{n} = [a + a + (n - 1) d] + [a + d + a + (n - 2) d] + . .

2 S_{n} = [2 a + (n - 1) d] + [2 a + (n - 1) d] + . . n t e r m s

2 S_{n} = n [2 a + (n - 1) d] (. n t e r m s)

2 S_{n} = n [2 a + (n - 1) d]

S_{n} = \frac{n}{2} [2 a + (n - 1) d]