0
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.
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.
Oh, dude, the sum of the first 100 multiples of 3 is like, um, 15,150. You just gotta take the first multiple, which is 3, and the last one, which is 300, and then use the formula for the sum of an arithmetic series. Easy peasy, lemon squeezy.
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 !
Add your answer:
Earn +20 pts
What is the sum of the first 4 multiples of 5?
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
What is the sum of the first 10 multiples of 3?
The sum of the first 10 multiples of 3 is 165.
What is the sum of the first 25 multiples of 3?
975
What is the sum of the first 4 multiples of 5?
5 + 10+15+20=50 ans
What is the sum of the first 500 multiples of 3?
Oh, what a happy little question! To find the sum of the first 500 multiples of 3, we can use the formula for the sum of an arithmetic series: (n/2) * (first term + last term). In this case, the first term is 3 and the last term is 3 * 500. Plugging these values in, we get (500/2) * (3 + 1500) = 250 * 1503 = 375,750.
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 88 positive multiples of 3?
Sum_ap = ½ × number_of_terms × (first_term + last_term) For the first 88 multiples of three: number_of_terms = 88 first_term = 1 × 3 = 3 last_term = 88 × 3 = 264 → Sum = ½ × 88 × (3 + 264) = 44 × 267 = 11748
What is the sum of the first 3 multiples of4?
4 + 8 + 12 = 24
What is the sum of the first 3 multiples of 10?
10+20+30 = 60
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
What is the sum of all the multiples of 3 and 5 below 500?
The sum of all multiples of 3 below 500 is the sum of 3 + 6 + ... + 498 = 41583The sum of all multiples of 5 below 500 is the sum of 5 + 10 + ... + 495 = 24750Depending upon the interpretation of "of 3 and 5", the answer is one of:The sum of all multiples of both 3 and 5 below 500 is the sum of 15 + 30 + ... + 495 = 8415The sum of all multiples of 3 below 500 and all multiples of 5 below 500 is 41583 + 24750 = 66333Since the multiples of both 3 and 5 (that is 15, 30, ...) have been counted twice - once in the multiples of 3 and once in the multiples of 5 - the sum of all multiples of 3 or 5 or both below 500 is 41583 + 24750 - 8415 = 57918I have emphasised the word below with regard to 500 since 500 is a multiple of 5, but 500 is not below (less than) itself, that is the multiples are of the numbers 1, 2, 3, ..., 499. If the question was intended to mean less than or equal to 500, add an extra 500 to the multiples of 5 above, making the sum of the multiples of 5 be 25250 and the final sums (1) 8415, (2) 66833, (3) 58418.
