Question

find the sum of all 3 digit number which are divisible of three​

Answer 1

$\huge \mathtt{ \fbox{Solution : }}$

Given ,

The AP is 102 , 105 , .... , 999

Here ,

First term (a) = 102

Common difference (d) = 3

Last term (an/l) ) = 999

We know that , the first nth term of an AP is given by

An = a + (n - 1)d

Substitute the known values , we get

➡999 = 102 + (n - 1)3

➡897 = (n - 1)3

➡299 = (n - 1)

➡n = 300

Also we know that , the sum of first n terms of an AP is given by

$\large \mathtt{ \fbox{S = \frac{n}{2}(a + l) }}$

Thus ,

$\sf \mapsto S = \frac{300}{2} (3 + 999) \\ \\\sf \mapsto S =150 \times 1002 \\ \\ \sf \mapsto S =150300$

Hence , the sum of all 3 digit number which are divisible of three is 150300

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Answer 2

Smallest 3 digit number divisible by 3 , a=102

Largest 3 digit number divisible by 3 , tn=999

Common difference(d)=3

Let the total number of three digit number be 'n'

Therefore,

$t_{n} = a + (n - 1)d$

substituting the values

$999 = 102 + (n - 1)3 \\ \\ 999 - 102 = (n - 1)3 \\ \\ 897 = (n - 1)3 \\ \\ (n - 1) = \frac{897}{3} \\ \\ n - 1 = 299 \\ \\ n = 300$

$s_{n} = \frac{300}{2} (2 \times 102 + (300 - 1)3) \\ \\ s_{n} = 150(204 + (299)3 ) \\ \\ s_{n} = 150(204 + 897) \\ \\ s_{n} = 150(1101) \\ \\ s_{n} = 165150$

By the formula

$s_{n} = \frac{n}{2} (2a + (n - 1)d)$