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Arithmetic Progression

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by sanju09 » Wed Sep 18, 2013 5:07 am
mgm wrote:Just wondering , what is the best way to solve the following :

What is the sum of sum of odd numbers between 50 and 100 and sum of even numbers between 200 and 300 ?
The features of a series that contains equally spaced terms:

"¢ The average (arithmetic mean) of the series is just the average of its first and last terms.

"¢ The number of terms in the series is one more than the ratio in the difference between last and first term to the equal space size.

"¢ Sum of the series is half the product of number of terms and sum of first and last terms.

With that information, the number of terms in the series 51, 53, 55...99 is 1 more than (99 - 51)/2 or 25, and sum of this series is ½ (25) (51 + 99), which is [spoiler]1875[/spoiler].

Similarly, the number of terms in the series 202, 204, 206...298 is 1 more than (298 - 202)/2 or 49, and sum of this series is ½ (49) (202 + 298), which is [spoiler]12250[/spoiler].

And, finally, sum of the two sums is [spoiler]1875 + 12250 = 14125[/spoiler].
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by theCodeToGMAT » Wed Sep 18, 2013 5:39 am
Use Arithmetic Progression

Sum of ODD Numbers from 50 to 100,
i.e. 51, 53, ... 99

S = n ( 2a + (n-1)d)/2
= 25 ( 2 x 51 + 24 x 2)/2
= 25 x 150 /2
= 1875

Or,

Sum of EVEN Numbers from 200 to 300,
i.e. 202, 204... 298

S = n ( 2a + (n-1)d)/2
= 49 ( 2 x 202 + 48 x 2 )/2
= 49 ( 500 ) /2
= 12250

Total Sum = 1875 + 12250 = 14125
mgm wrote:Just wondering , what is the best way to solve the following :

What is the sum of sum of odd numbers between 50 and 100 and sum of even numbers between 200 and 300 ?
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by Brent@GMATPrepNow » Wed Sep 18, 2013 5:58 am
Another approach for sum of ODD Numbers from 50 to 100,

We want 51+53+55+....97+99

There are 50 numbers from 51 to 100 inclusive (since 100-51+1 = 50)
Half of these numbers are odd and half are even, so we know there are 25 terms in the sum 51+53+55+....97+99

For "fun," let's evaluate (51+53+55+....97+99) + (51+53+55+....97+99), but we'll do so in a useful way:

....51 + 53 + 55 +...+ 97 + 99
.+ 99 + 97 + 95 +...+ 53 + 51
...150+ 150+ 150+...+ 150 + 150

How many 150's do we have in the new sum? There are 25 altogether.
25x150 = 3750

IMPORTANT: Since we've added every term twice, we need to take 3750 and divide by 2 to get 1875

Cheers,
Brent
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by theCodeToGMAT » Wed Sep 18, 2013 5:59 am
Another Method:

Sum of First n Even Number: n(n+1)
Sum of First n Odd Number : n^2

Sum of ODD Number between 50 to 100 = (Sum of ODD numbers 0 to 100) - (sum of 0 to 50)
 = 50^2 - 25^2
= (50+25) x ( 50-25)
= 75 x 25 = 1875


Sum of EVEN Number between 200 to 300 = (Sum of EVEN numbers 0 to 300) - (sum of 0 to 200)
 = (149(149+1)) - (100(100+1))
= (149 x 150 - 100 x 101)
= 50 ( 149 x 3 - 2 x 101)
= 50 (245) = 12250
Total Sum = 1875 + 12250 = 14125

mgm wrote:Just wondering , what is the best way to solve the following :

What is the sum of sum of odd numbers between 50 and 100 and sum of even numbers between 200 and 300 ?
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