Question

Find the sum of all natural from 1 to300 divisible by 3​

Answer 1

Answer

15,150

$\rule{200}2$

Solution

The first natural number divisible by 3 is '3' itself.

The last number between 1 and 300 that is divisible by 3 would be '300'.

Hence, the numbers between 1 and 300 divisible by 3 would be written as -

3, 6, 9....... 300.

We can see that it is forming an A.P. with first term (a) = 3

common difference (d) = 3

Last term (l) = 300.

Now, applying the formula of general term $\sf T_n$, we can find the number of terms in the above A.P.

→ a + (n - 1)d = 300

→ 3 + (n - 1)3 = 300

→ (n - 1)3 = 300 - 3

→ n - 1 = 297/3

→ n - 1 = 99

→ n = 100

Now, sum of the A.P. = $\sf\frac{n(a + l)}{2}$

Where, a is the first term, n is the number of terms and l is the last term.

Then, the sum of A.P. would be

100(3 + 300)/2

50(303)

= 15,150

Hence, the sum of natural numbers between 1 and 300 which are divisible by 3 is 15,150.

Answer 2

Given,
D=3
A=3
An=300

We know,
An=A+(n-1)D
300=3+(n-1)(3)
100=3+(n-1)
97=n-1
n=97+1=98

We know, Sn=n/2(A+An)
= 98/2(3+300)
=49(303)
= 14,847


Hope my answer helped you....
Please mark me brilliant....