Derive the formula for the sum of n terms of an ap
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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]
-------------⚠---------------------------
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
⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠
Hope it helps you !!!
#Rajukumar111@@@
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]
-------------⚠---------------------------
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
⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠⚠
Hope it helps you !!!
#Rajukumar111@@@
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