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Let the first number be 2n+1,If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.
Here's another approach:ahinye wrote:If S is a series of x consecutive odd integers and Z is the larget integer in S, what is the smallest integer in S.
1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)
How can I solve this algebraically?
It is a very simple question of APahinye wrote:If S is a series of x consecutive odd integers and Z is the largest integer in S, what is the smallest integer in S.
1) Z-2x
2)Z-x+1
3)Z-2(x-1)
4)x-6+Z
5)Z- (x/2)
How can I solve this algebraically?
I don't think plugin is at all required here. Just go with simple concept of AP and u will get ans in seconds.. Though plug in really helps a lot many a times..gmatmachoman wrote:Agreed .But plugging in numbers are pretty easy.
take ,5,7,9.....
HereZ=9,x=3
We know that our smallest number is 5.
Plug in the values of Z & X.
1.Z-2x
--->9- 2*3=3( Not 5, which is required)
2.Z-x+1
---> 9-3+1=7(Not the answer we want)
3.Z-2(x-1)
--->9-2(3-1)=5(Bingo! we got answer smallest integer)
Well,I would say it should be decided upon seeing the question which technique to use:-In some cases that I have personally found,solving by the formal proper approach takes too much time and even the calculation process is very cumbersome.ahinye wrote:Thank you for this explanation. Looks easier when it has been layed out logically. Was going round in circles and had to post something. thanks for all the responses thus far. I prefer to solve with algebra than plugging numbers where possible
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