Question

3+5+7+............. +401 Please find the sum

Answer 1

Answer:

it is in AP

so d= 2 ; a=3 ; An term = 401

An^th = a+(n-1)d

401= 3+(n-1)2

398 = (n+1)2

n= 198

sum = n/2(a+An^th)

= 198/2(3+401) =39996

Answer 2

Answer:

it is in AP

so d= 2 ; a=3 ; An term = 401

An^th = a+(n-1)d

401= 3+(n-1)2

398 = (n+1)2

n= 198

sum = n/2(a+An^th)

= 198/2(3+401) =39996