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# The sume of the first 100 positive multiples of 4 is _ more than the sum of the first 100 positive multiples of 3.

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The sume of the first 100 positive multiples of 4 is _ more than the sum of the first 100 positive multiples of 3.

A. 100

B. 400

C. 1200

D. 5050

Feb 6, 2019

#1
+4325
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Compare the sums for a second:

4+8+12+16+20+24

3+6+9+12+15+18

Sum of the first 100 positive integers, so 5050 or (D).

Feb 6, 2019
#3
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That is correct: sum of 1 to 100 =5050.

Guest Feb 6, 2019
#2
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The arithmetic sum formula shows that:

$$\text{sum} = 100(\frac{4+100}{2}) = 5200$$

$$\text{sum} = 100(\frac{3+100}{2}) = 5150$$

The difference is 5200 - 5150 = 50.

Feb 6, 2019
#4
+104804
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tertre is correct

4 + 8 + 12 + 16 +.....+  400

3 + 6  + 9  + 12 +.....+  300

The difference in the series is

1 + 2 + 3 + 4 + ....+ 100   =  the sum of the first 100 positive integers =   100 (101) / 2  =  50 * 101 = 5050

Feb 6, 2019