Question

find the sum of all natural number between 200 and 500 which are divisible by 4 ​

Answer 1

d=4

a=204

an=496

We know that n =an-a
——- +1
d


ie , 496-204/4 +1

292/4+1

73+1

=74. Terms


Now, n=74


From above given factors Sn=n/2(a+an)


ie,,,,, 74/2(204+496)


74/2(700)

74*350

=25900




Ans =25900




Please mark it as .. if u think

Answer 2

A.P. for given conditions is..204,208,212,....496.

Here,

a=204, d=4 & tn=496

Sol.

tn=a+(n-1)d

496=204+(n-1)4

496-204=4n-4

292+4 =4n

296/4 =n

n =74

Now,

Sn=n/2[t1+tn]

S74= 74/2[204+496]

S74= 37(700)

S74= 25,900,,