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Expert replies
by klaud » Thu Mar 15, 2012 9:33 am
If the first number of a number list is 4 and each of any successive number is 2 more than the precedent, which is the number at 77th place?

How i solve this question?
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Source: — Problem Solving |

by Mike@Magoosh » Thu Mar 15, 2012 3:31 pm
Hi, there. I'm happy to help with this. :)

We want to go to the 77th place on the list. One important idea is that if we take a "step" from one place on the list to the next, then we need 76 steps to get from place #1 to place #77.

Then reason is --- think about it --- if you start at place #1, and take three steps, those steps move you 1 ---> 2, then 2 ---> 3, then 3 ---> 4, so three steps move you to place #4. Nine steps would move you from place #1 to place #10. If you start at place #1 and want to get to place #n, you need (n-1) steps.

So, we start with a value of 4 at place #1. To get to the 77th place, we take 77-1 = 76 steps, each of which has a length of 2, so together, that's an increase of 2*76 = 152. Now, add that increase to the initial amount, and we get 4 + 152 = 156. That's the value at the 77th place.

A more formal approach. This is called an arithmetic sequence, where each successive term is the previous term plus a fixed difference. The formula for the nth term of an arithmetic sequence is

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

where a(1) is the initial term, and d is the fixed difference between terms.

Does that make sense? Let me know if you have any questions.

Mike :)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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