Question

find the sum of numders from 0 to 200 which are divisible by both 3 and 4​

Answer 1

Answer:

first no. which is divisible by both 3 & 4 is 12

So, a=12 and did also equal to 12

last term which is divisible by 12 is 192 within the range of 0-200

Tn=a+(n-1)d

192=12+(n-1)12

192-12=(n-1)12

180=(n-1)12

180/12=n-1

15=n-1

n=16

Sn=n/2(a+l)

S16=16/2(12+192)

S16=8*204

Ans=1632