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by bmanavi » Tue Jun 08, 2010 3:40 pm
Can anyone explain to me how to do the following question? I dont understand the OG assumptions and the process used to yield a result....

Question: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is th sum of all the even integers between 99 & 301...

THANKS!
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by krazy800 » Tue Jun 08, 2010 3:54 pm
bmanavi wrote:Can anyone explain to me how to do the following question? I dont understand the OG assumptions and the process used to yield a result....

Question: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is th sum of all the even integers between 99 & 301...

THANKS!
The even numbers are between 99 and 301 are ... 100, 102, 104....300

n =( (300-100)/2) + 1 (we are adding 1 because we are including 100)

n=101

n(n+1)/2 = 101*56 = 5656

HTH!!
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by eloka » Tue Jun 08, 2010 5:40 pm
Ok I am confused ^^

301 - 99 = 212 + 1 = 213/2 is not divisible at all....Don't you need to add with 1 first before we divide the "numbers" with 2??

301 - 99 = 212
212 + 1 = 213

somebody help me.
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by krazy800 » Tue Jun 08, 2010 5:51 pm
eloka wrote:Ok I am confused ^^

301 - 99 = 212 + 1 = 213/2 is not divisible at all....Don't you need to add with 1 first before we divide the "numbers" with 2??

301 - 99 = 212
212 + 1 = 213

somebody help me.

eloka:

The question stem asks for sum of even numbers between 99 and 301

so the even number series starts from 100 to 300

Total even numbers between 99 and 301 i.e. n= ((300-100)/2) + 1 (add one to take into include 100)

n= 101

now sum of numbers = n(n+1)/2 = 101*102/2 = 5656


To further explain consider the following number series

1, 2, 3, 4

how many integers do we have including 1 and 4 ..... (4-3)+ 1 =4 right ( remember This is applicable only if the question system asks to find out number of integers between 1 and 4 inclusive of 1 and 4)

If the question system asks you to find out the numbers bewteen 1 and 4 i.e.exlusive we just do 4-1=3 integers

Hope I am not confusing you

let me know if you need more help!
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by muralithe1 » Tue Jun 08, 2010 6:20 pm
Well..The answer 5656 is wrong...


But ofcource the number of elements is 101...but of the sum of those 101 elements is not 5656..

It should be derived by the below formula.

"
To find the sum of a certain number of terms of an arithmetic sequence:Sn = n(a1+an)
2
where Sn is the sum of n terms (nth partial sum), a1 is the first term, an is the nth term."

Here a1= 100,a2=300,n=101

So Sn=101(100+300)/2=20200.

Please correct me if i am wrong.
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by Patrick_GMATFix » Tue Jun 08, 2010 6:25 pm
This is #157 in OG12 with OA = B. The formula given is not needed to calculate the answer. Instead, it would be more productive to learn how to deal with sequences in which the difference between consecutive terms is always the same (Arithmetic sequences). In general these sequences have the following properties: Average = median = average of max and min. Sum = avg * # of terms (this is a variation of the average formula avg=sum/#of terms).

2 solutions, a 30-second hack and take-away lessons are attached.

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-Patrick
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by krazy800 » Tue Jun 08, 2010 6:34 pm
muralithe1 wrote:Well..The answer 5656 is wrong...


But ofcource the number of elements is 101...but of the sum of those 101 elements is not 5656..

It should be derived by the below formula.

"
To find the sum of a certain number of terms of an arithmetic sequence:Sn = n(a1+an)
2
where Sn is the sum of n terms (nth partial sum), a1 is the first term, an is the nth term."

Here a1= 100,a2=300,n=101

So Sn=101(100+300)/2=20200.

Please correct me if i am wrong.

u r right! I have wrongly used formula from the above post. the formula n(n+1)/2 is used to calculate sum of first n natural numbers.

Thanks for correcting!!!!
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by mepinoargote » Thu Jun 17, 2010 1:50 pm
Thanks a lot Patrick_GMATFix! great explanation about arithmetic sequences. Helped a lot.
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