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Re: For all even integers n, h(n) is defined to be the sum of the even
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31 Aug 2016, 05:13

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mystiquethinker wrote:

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

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) = [(2+18)/2]*9 = 90

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

h(18)/h(10) = (90) / (30) = 3
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Re: For all even integers n, h(n) is defined to be the sum of the even
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31 Aug 2016, 07:48

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mystiquethinker wrote:

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

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 [yes, I'm aware that the fraction evaluates to be exactly 2, but we could actually be quite aggressive in our estimation and still reach the correct answer] ≈ 3

Re: For all even integers n, h(n) is defined to be the sum of the even
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22 Aug 2017, 10:46

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

Re: For all even integers n, h(n) is defined to be the sum of the even
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24 Aug 2017, 16:05

mystiquethinker wrote:

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

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):

Re: For all even integers n, h(n) is defined to be the sum of the even
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22 Mar 2018, 03:31

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