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The solution to the given problem can be obtained by sum formula of arithmetic progression.

In arithmetic progression difference of two consecutive terms is constant. The multiples of any whole number(in sequence) form an arithmetic progression.

The first multiple of 3 is 3 and the 100th multiple is 300.

3, 6, 9, 12,... 300. There are 100 terms.

The sum 3 + 6 + 9 + 12 + ... + 300 can be obtained by applying by sum formula for arithmetic progression.

Sum = (N/2)(First term + Last term)

where N is number of terms which in this case is 100.

First term = 3; Last term = 300.

Sum = (100/2)(3 + 300) = 50 x 303 = 15150.

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10y ago
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Swasthika Priyadarsh...

Lvl 2
4mo ago

Its easy to see that there are 33 multiples of 3 between 1 and 100 as 3*33=99 which is the nearest multiple of 3 less than 100. The A.P. will look like:

3, 6, 9, 12,…………., 96, 99

So, we have the first term as 3, common difference as 3 and number of terms as 33, just put them into the formula S= n(2a + (n-1)d)/2 where n=number of terms, a=first term and d=common difference, to get the answer as 1683.

In case you want to know how to arrive at this formula, put it in the comments and i’ll give you the derivation but i suggest you try it on your own, its really easy.

HAPPY LEARNING !

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Anonymous

Lvl 1
4y ago

the sum of the first 100 multiples of 3 is 15150

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Q: What is the sum of the first 100 multiples of 3?
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