Question

Find the sum of all natural numbers from 1 to 200
i) which are divisible by 5
ii) which are divisible by 3.

Answer 1

1) the natural number from 1 to 200 divisible by 5 are 10,15,20.......200.
the above sequence is an A.P.
a=10, d=15-10=5
let number of terms in A.P. be in
then tn=200
tn=a+(n-1)d
200=10+(n-1)(5)
200-10=(n-1)(5)
190=(n-1)5
190/5=n-1
18=n-1
18+1=n
19=n
n=19

Answer 2

i) 1st number =5
2nd = 10
last = 200
they are in AP
AP=5,10,15,20........200
an=a+(n-1) d
200=5+(n-1)5
195=(n-1)5
39=n-1
40=n
$s40 = \frac{40}{2} (5 + 200) \\ = 20(205) \\ \huge\bold{4100}$
ii)here also an AP.
3,6,9.....198
an=a+(n-1)d
198=3+(n-1) 3
195/3=n-1
65=n-1
66=n
$s66 = \frac{66}{2} (3 + 198) \\ = 33 \times 201 \\ \huge\bold{6633}$