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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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Jordan
RossOh, what a delightful question! To find the sum of the first 100 multiples of 3, we can use a formula. The sum is calculated by multiplying the total number of multiples (100) by the average of the first and last term (3 and 300). So, the sum would be 100 multiplied by 153, which equals 15,300. Isn't that just lovely?
To find the sum of the first 100 multiples of 3, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term. In this case, a1 = 3 and an = 3 * 100 = 300. Plugging these values into the formula, we get: Sn = 100/2 * (3 + 300) = 50 * 303 = 15,150. Therefore, the sum of the first 100 multiples of 3 is 15,150.
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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5 + 10+15+20=50 ans
What is the sum of all the multiples of 3 between 100 and 200?
this isn't a place to do your math homework.
What is the sum of the first 25 multiples of 4?
Sum of first 25 multiples of 44+8+12....100taking 4 common4(1+2+3+4....+25) = 4*325 =1300
How many multiples of 3 are there between 100 and 500?
There are 166 multiples in 500 (500/3=166r2)and 33 multiples in100(100/3=33 r1) if you subtract the later from the first.... 166-33=133
What is the sum of the first 15 multiples of 3?
3+6+9...+45taking 3 common3(1+2+...+15) = 360
