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by kmittal82 » Thu Sep 09, 2010 7:02 am
Term at position n = 5 + (n-1)x16

Thus, 123 term = 5 + (123-1)x16 = 5 + 1952 = 1957
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by kvcpk » Thu Sep 09, 2010 8:01 am
pradeepkaushal9518 wrote:if 5 is the first term of a series and 16 is added to each succesive term then what is the ones digit of 123 rd term
Mittal is right. Just in case you are not familiar with Arithmetic progression formulas, you can check the pattern.

5,21,37,53,69,85...

So, after every 5 numbers, the last digits repeat.

123 divided by 5 gives reminder 3.

Hence pick the third entry from the list.
37 has last digit 7.

Hence 7 is the answer.

Hope this helps!!
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by beatthegmatinsept » Thu Sep 09, 2010 8:03 am
hey guys - where did you get this formula from? I have gone thru MGMAT books and don't remember seeing this formula.. Looks like I just learnt something new! Very quick way of solving this problem. :)
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by kvcpk » Thu Sep 09, 2010 8:10 am
beatthegmatinsept wrote:hey guys - where did you get this formula from? I have gone thru MGMAT books and don't remember seeing this formula.. Looks like I just learnt something new! Very quick way of solving this problem. :)
Its actually pretty simple.


The nth term of an Arithmetic progression, a,a+d,a+2d,..... is given by a+(n-1)d

Sum of n-terms is given by n(2a+(n-1)d)/2

Very useful and Handy formulas.
Add them to your flashcards!!
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by kmittal82 » Thu Sep 09, 2010 8:20 am
AFAIK, AP is not tested on the GMAT.

The original question posted doesn't need the "AP formula" per se, more common sense than anything :)
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by beatthegmatinsept » Thu Sep 09, 2010 8:25 am
That's what I would have done, add 16 to each successive term and look for the trend. But considering how precious time is on the GMAT, the formula seems like it might save a few seconds.
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