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there must be a formula for this ?

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by Atult718 » Tue Apr 17, 2007 1:30 am
I am sure there would be an easier method or a formula to solve the problem below?

Q: If A equals the sum of the even integers from 2-20, inclusive, and B equals the sum of the odd integers from 1-19, inclusive, what is the value of a-b ?

A) 1
b) 10
C) 19
D) 20
E) 21
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by rajesh_ctm » Tue Apr 17, 2007 4:11 pm
There is an easy method. Formula might be more time-taking.

a = 2+4+6+......20
b = 1+3+5+......19

subtract b from a, it will be (2-1)+(4-3)+(6-5)+.......(20-19)
so a-b = 1+1+1+......1 = 10
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by Atult718 » Tue Apr 17, 2007 11:02 pm
right .. this way works... but i mean isn't this method too time consuming to be used on gmat?... it is almost same as adding all the numbers and then finding out the difference ?
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use AP formula

by crossingfingers » Wed Apr 18, 2007 7:17 am
use Arithmetic Progression formula:

Sum = n/2[a +l] where n = number of integers; a= first number; b=last number in the sequence

solve for the 2 groups and subtract...the answer is 10.
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by rajeshvellanki » Fri Apr 20, 2007 7:17 am
a(n)=a1+(n-1)d
S=(n/2)(a1+an)

d=difference between the consecutive numbers

for the first series a=2+4+6+...20

d=2
a1=2
an=20

a(n)=a1+(n-1)d

20=2+(n-1)2
=>n=10

S=(10/2)(2+20)
=>S(a)=110

same for b
=>n=10
and
S=(10/2)(1+19)
=>S(b)=100

S(a)-S(b)=110-100=10

You should be able to do most of the math in your head.

R
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by Stacey Koprince » Mon Apr 30, 2007 1:08 pm
If you have consecutive integers (or consec even / consec odd), you can find the sum by finding:
1) the average of the first and last terms, and
2) the # of terms

and then multiplying those two together.
So, for even from 2-20
1) avg = 22/2 = 11
2) # = (20/2) - (2/2) + 1 = 10-1+1 = 10 (Note: need to divide by 2 b/c only counting the even terms)
multiply to get 110

For odd from 1-19
1) avg = 20/2 = 10
2) # = 10 (can tell b/c there were 10 terms for 2 to 20 - same as 1 to 19)
multiply to get 100

difference = 10
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by parore26 » Wed May 02, 2007 6:05 am
Wow! what a great thread. I was certain when I saw this question there was only one way to solve it: Using the formula for adding consecutive integers. I liked Rajesh's solution. Cheers guys, I'll remember the two new tricks I learnt today to deal with arithmetic progressions...
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