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प्रश्न
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
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उत्तर
In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.
So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.
Also, all these terms will form an A.P. with the common difference of 3.
So here,
First term (a) = 3
Last term (l) = 99
Common difference (d) = 3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
`a_n = a + (n - 1)d`
So, for the last term,
99 = 3 + (n - 1)3
99 = 3 + 3n - 3
99 = 3n
Further simplifying
`n = 99/3`
n = 33
Now, using the formula for the sum of n terms,
`S_n = 33/2 [2(3) + (33 - 1)3]`
`= 33/2 [6 + (32)3]`
`= 33/2 (6 + 96)`
`= (33(102))/2`
On further simplification, we get,
`S_n = 33(51)`
= 1683
Therefore, the sum of all the multiples of 3 lying between 1 and 100 is `S_n = 1683`
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