Question

find the sum of integers from 1 to 100 which are divisible by 2 or 3​

Answer 1

To find the sum of the integers ranging from 1 to 100 which are divisible by 2 or 3

Here,

I would like to remind you the divisibility rule of 6↓↓↓

i.e.,a number which is divisible by both 2 and 3 is also divisible by 6

Implies,

We have to find out the sum of the integers from 1 to 100 which are divisible by 6.

Consider ‘a’ and ‘d’ to be the first term and common difference of the following A.P

★The A.P would be,

6,12,18,24,........96

Now,

First term=a=6

Common difference=d=6

Last term=l=96

We have to first find the no.of terms

Thus,

l=a+(n-1)d

→96=6+(n-1)6

→6n=96

→n=6

★Sum of the terms:

Using a=6,l=96 and n=16

$\sf{ \frac{n}{2}(a + l) } \\ \\ = \sf{ \frac{16}{2}(96 + 6) } \\ \\ \sf{ = 102 \times 8} \\ \\ \underline{ \sf{ = 816}}$

The sum of the terms is 816

Answer 2

Given question is to find the sum of the integers from 1 to 100 which are divisible by 6.

Consider ‘a’ as first term and 'd’ as common difference of the following A.P

Here , The A.P is :

6,12,18,24,........96

Now,

a=6

d=6

l=96

We have to first find the no.of terms

Thus,

l=a+(n-1)d

96=6+(n-1)6

6n=96

n=6

Now , Sum of the terms:

Using a=6 ,l=96 and n=16

Use n/2(a+I) formula and substitute

Answer is 816

Therefore,

The sum of the terms is 816