For all even integers n, h(n) is defined to be the sum of the even

Question

For all even integers n, h(n) is defined to be the sum of the even integers between 2 and n, inclusive. What is the value of h(18)/h(10) ?

(A) 1.8
(B) 3
(C) 6
(D) 18
(E) 60

GMATinsight · Accepted Answer

CONCEPT : When terms are in Arithmetic Progression (A.P.) i.e. terms are equally spaced then

Mean = Median =(First+Last)/2

and Sum = Mean*Number of terms

h(18) =  *9 = 90

h(10) = (2+10)/2]*5 = 30

h(18)/h(10) = (90) / (30) = 3

BrentGMATPrepNow · Answer

We can also apply a nice  rule  that says   (a + b)/c = a/c + b/c   AND use some  estimation , since the answer choices are  nicely spread apart .

h(18)/h(10) = ( 2+4+6+8+10 + 12+14+16+18 )/( 2+4+6+8+10 )
= ( 2+4+6+8+10 )/( 2+4+6+8+10 ) + ( 12+14+16+18 )/( 2+4+6+8+10 )
= 1 + ( 12+14+16+18 )/( 2+4+6+8+10 )
= 1 + some value around 2    
≈ 3

Answer:  B

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rhine29388 · Answer

sum of n terms can be represented by the formula =  \frac{n*(n+1)}{2}

h(10) :- 2+4+6+8+10 = 2(1+2+3+4+5) = 2 *  \frac{5*6}{2}  = 30

h(18) :- 2+4+6+8+10+12+14+16+18 = 2(1+2+3+4+5+6+7+8+9) =  \frac{2 * 9 *10}{2}  = 90

\frac{h(18)}{h(10)}  =  \frac{90}{30}  =3

correct answer - B

Esguitar · Answer

Hi,
Could I apply the following?

h(18)/h(10)  simplified is  h(9)/h(5) , so the even numbers are  (2+4+6+8)/(2+4)  = 3

Thanks

NamVu1990 · Answer

h(10) = 2+4+6+8+10 = 30
h(5) = 2+4 =6
so h(10) can't = 2 x h(5), you can't simplify like that.

sobby · Answer

The number are in AP 
we can use n/2(2a + (n-1)d)
we will get sum of n even terms ---( a = 2 , d = 2) =n(n+1) 
Now 2 to 18 we have 9 terms
2 to 10 we have 5 terms
9*10 / 5*6
we will get 3

ScottTargetTestPrep · Answer

First, notice that h(18) will be the sum of the even integers between 2 and 18 inclusive. We can thus calculate h(18) using the formula sum = avg x quantity.

First, find the average of this evenly spaced set of numbers:

avg = (18 + 2)/2 = 10

Now calculate how many numbers are in this set:

quantity = (18 - 2)/2 + 1 = 9

Now we use the formula sum = avg x quantity:

sum = (10)(9)

sum = 90 = h(18)

Next we can determine h(10) using the same procedure we used to calculate h(18):

avg = (10 + 2)/2 = 6

quantity = (10 - 2)/2 + 1 = 5

sum = 30

Thus, h(18)/h(10) = 90/30 = 3

Answer: B

NDND · Answer

Hi Everyone,

I used a different way here

A modified version of  \frac{(n*(n+1))}{2}  would be  \frac{(n*(n+2))}{4}

h(18)=\frac{(18)(20)}{4} = \frac{360}{4}= 90

h(10)= \frac{(10)(12)}{4} = \frac{120}{4}= 30

\frac{h(18)}{h(10)}= \frac{90}{30}= 3  ===> B

gerikso · Answer

We can also apply a nice  rule  that says   (a + b)/c = a/c + b/c   AND use some  estimation , since the answer choices are  nicely spread apart .

h(18)/h(10) = ( 2+4+6+8+10 + 12+14+16+18 )/( 2+4+6+8+10 )
= ( 2+4+6+8+10 )/( 2+4+6+8+10 ) + ( 12+14+16+18 )/( 2+4+6+8+10 )
= 1 + ( 12+14+16+18 )/( 2+4+6+8+10 )
= 1 + some value around 2    
≈ 3 
_______________________________________________________________________

Can I use another nice rule   ab/cd   for this? 
The answer is also a close approximation to B. If h(18)/h(10) = h/h * 18/10 = 9/5. 
In the numerator, the even terms would be 2+4+6+8=20. 
In the denominator, the even terms would be 2+4=6. 
Therefore, 20/6=~3.
Please explain why "yes"/"no". Thank you!

Kimberly77 · Answer

Hi Brent  BrentGMATPrepNow , is this way of calculation correct too?

n∗(n+1)/2 = (18 * 19)/2 = 171
n∗(n+1)/2 = (10 * 11)/2 = 55

h(18)/h(10) = 171/55 = 34/11 > 30/10 approximately 3

BrentGMATPrepNow · Answer

Given: h(n) is defined to be the sum of the   even   integers between 2 and n, inclusive

So, h(18) = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90
And, h(10) = 2 + 4 + 6 + 8 + 10 = 30

The formula that you are using is for the sum of all integers from 1 to n inclusive (not just the even integers).

So, it's merely a coincidence that you got the correct answer.

Kimberly77 · Answer

Thanks BrentGMATPrepNow for clarification. Noted thanks

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For all even integers n, h(n) is defined to be the sum of the even

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