Question

Derive the formula for the sum of n terms of an A.P

Answer 1

Hey !!!

if 1st term of an Ap is a

then , second term t2 = a+(2-1)d = a+ d

similarly , t3 = a+ 2d

and so ,nth term of an AP = a+(n-1)d

let S denote the sum of the first m term of The AP ,

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

Rewriting the terms inverse order

we get ,

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

now , on adding (1 ) and (2) by term wise

we get ,

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

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n times

or, 2S = n[2a+(n-1)d ] •°• there is nth terms

or, S {n/2(2a+(n-1)d]

So , the sum of the first n term of ap AP is given by

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

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Hope it helps you !!!

Mark as brainliest please!!!

Answer 2

Answer:

hi

Step-by-step explanation:

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