Question 194516: what is sum of all multiples of 6 that are strictly between 3138 and 51804?
please help i my exam is 05/07/09 so it is urgent please help!!!!!!
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Often times we can use simple numbers to define the operations needed to solve actual.
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lets use 2,4,6, ,,,,sum is 12
average (mean) is 12/3 = 4,,, if we only use end points ,,, avg = (2+6)/2=4
number of multiples = no of spaces +1,,, or ( 6-2)/2 =2 spaces +1 =3 numbers
sum = avg * num = 4*3 =12
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This leads to the A*N=T formula,,, average * number =total
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from above we generalize, average = (min + max) / 2
or ( 51804 + 3138 ) / 2 = 27471
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number of spaces = (max - min ) / size of interval = ( 51804 - 3138 ) / 6 = 8111
number of numbers = 8111 +1 = 8112
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Sum is avg * no = 27471 * 8112 = 222,844,752
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An alternate solution would be to use the arithmetic series
a(n) = a(1) + (n-1) d
51804 =3138 + (n-1) 6
48666/6=n-1
8111=n-1
n=8112
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Sum formula is s={ (a(1) +a(n) )/2 } n = {(51804 +3138)/2} 8112 = 27471 *8112 = 222,844,752
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check is ok
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