Question

Find the sum of all natural numbers between 1 and 100, which are divisible by 3

Answer 1

Hint: We have to expand the given natural numbers in series and solve using arithmetic progression and find the sum of the natural numbers.

Complete step-by-step answer:

We need to find the sum of all natural numbers between 1 and 100 that are divisible by 3
So all natural numbers from 1 and 100 that are divisible by 3 is
3, 6, 9, 12 ……….99
Clearly this forms an AP with first term a = 3 and common difference $d = {a_2} - {a_1} = 6 - 3 = 3$
Now nth term of an AP is ${a_n} = a + \left( {n - 1} \right)d$ here ${a_n} = 99$
So substituting values to find n we have
99 = 3 + (n – 1) 3
This gives 3n = 99 or n = 33
Hence there are in total 33 terms in this AP
So formula for ${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ where l is the last term
Hence ${S_n} = \dfrac{{33}}{2}\left( {3 + 99} \right)$
Therefore our ${S_n} = 1683$
Hence the sum of all natural number between 1 and 100 which are divisible by 3 is 1683

Note: In such questions always write the series it eventually be an AP or sometimes even an GP then use the respective formula to reach upto the answer.