|
Question 27695: what are three consecutive even intergers whose sum is 114?
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! what are three consecutive even intergers whose sum is 114?
Any two consecutive even integers differ by 2
Let the required 3 consecutive even integers be
(a-2), a and (a+2)
By data the sum of the 3 numbers is 114
Therefore we have
[(a-2) + a + (a+2)] = 114
(a-2+a+a+2)=114
which implies (a+a+a) = 114
(using additive commutativity and associativity
and the additive inverse law:-2+2=0)
That is 3a = 114
Therefore a = 114/3 = 38.
Hence the three consecutive even integers are (38-2), 38 and (38+2)
Answer: 36,38 and 40
Verification:
Of course 36,38 and 40 are all even (as each is a multiple of 2)
And their sum = 36 + 38 +40 = 114 which is correct.
One question? Why should we consider the required numbers in the above fashion?
Why not of the form a, (a+2), (a+4) ?
This form is ofcourse right and that is what we normally tend to proceed with.
There is nothing wrong with it.
The sum here is [a +(a+2) +(a+4)] = 114
After applying additive commutativity and associativity,we get
3a + 6 = 114
3a = 114-6
3a = 108
a = 108/3 = (36X3)/3 = 36
And here the numbers are 36, 36+2, 36+4 that is 36,38 and 40 of course
What is the advantage of the first method?
If we consider in the fashion:
(a-2), a and (a+2)
then the sum can be visualised mentally and the answer 3a= 114 flashes ina jiffy as a result and we immediately get the answer a =36. Cool! is n't it? That is the advantage.
|
|
|
| |
