Question

the sum of all natural numbers between 100 and 200 which are multiples of 3 is

Answer 1

first term is 102
last term is 198
no of terms is 
tn=a+(n-1)d
198=102+(n-1)3
96=3n-3
99=3n
n=33

sn=n/2(2a+(n-1)d)
     =33/2(2*102+32*3)
     =4950

Answer 2

Answer:

first term = 102. a= 102

last term. = 198

an= a+( n-1)d

198= 102+(n-1)3

198= 102+(3n-3)

198-102= 3n-3

96+3 = 3n

99/3=n

n= 33

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

sn= 33/2(2×102+102(33-1)3

sn=4950