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by MBA.Aspirant » Thu Jul 14, 2011 2:59 am
1) The general term for a series is given as an= a(n-1)-a(n-2). If a1=-5 and a2=4. What is the sum of the first 100 terms?

2) A series of terms is given as 3, 33, 333, 3333, ....... What is the hundred place of sum of first 10 terms of this series?
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by pemdas » Thu Jul 14, 2011 3:30 am
a(3)=a(2)-a(1)
a(1) a(2) a(3) -> a(3)=4-(-5)=9, a(4)=9-4=5, a(5)=5-9=-4, a(6)=-4-5=-9, A(7)=-9-(-4)=-5
-5,4,9,5,-4,-9,-5
the cycle repeats several times for the each six terms, thus 100/6={16 six terms from the cycle and four terms -> -5,4,9,5}

summing up (-5+4+9+5-4-9)=0 and multiplying 0 by 16 will return 0. Adding (-5+4+9+5)=13 should be the answer
MBA.Aspirant wrote:1) The general term for a series is given as an= a(n-1)-a(n-2). If a1=-5 and a2=4. What is the sum of the first 100 terms?
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by MBA.Aspirant » Thu Jul 14, 2011 3:38 am
Thanks pemdas. I got the same answer but the process was very tedious.

a3 = 4 - (-5)= 9

a4 = 9-4 = 5

a5 = 5-9 = -4

a6 = -4 -5 = -9

a7 = -9 - (-4) = -5

a8 = -5 - (-9) = 4

-5, 4, 9, 5, -4, -9, -5, 4, 9, 5

so as you said all 6 consecutive integers add to 0. so we have 16 times this will happen, leaving a97, a98, a99, and a100, which add up to 13.

for the 2nd one this is what I did:

3 +33+333+333+333 + 333 + 333+ 333+ 333 +333

adding for unit digit:

3* 10 = 30 or 0 carry 3

tenth:

3*9 = 27 +3 = 30 or 0 carry 3

hundred:

3*8= 24 +3 = 27

so hunderd digit is 7

700

so again a tedious approach.. is there a simpler one or these are just difficult qs?
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by GMATGuruNY » Thu Jul 14, 2011 3:44 am
MBA.Aspirant wrote:2) A series of terms is given as 3, 33, 333, 3333, ....... What is the hundred place of sum of first 10 terms of this series?
Of the 10 terms, all but the first two will have 333 as the last 3 digits:
333*10 = 3330.

To account for 33, subtract 300, since 333-300=33.
To account for 3, subtract 330, since 333-330=3.

3330-300-330 = 2700.

The hundreds digit is 7.
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by blackjack » Thu Jul 14, 2011 7:10 am
Hi Mitch, will it also be okay if do the following:
multiply 333*8=2664
and then add 33+3= 36 to it= 2700.

Somehow seems simpler and more obvious then subtracting 300 and 330.
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by GMATGuruNY » Thu Jul 14, 2011 10:09 am
blackjack wrote:Hi Mitch, will it also be okay if do the following:
multiply 333*8=2664
and then add 33+3= 36 to it= 2700.

Somehow seems simpler and more obvious then subtracting 300 and 330.
Your method is fine. I chose mine because it seems easier and quicker to multiply 333*10 than to multiply 333*8.
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by kitg » Thu Jul 14, 2011 7:40 pm
Answer to question 2:

asked thing can be found by
adding all the 3's in the unit's digit of the 10 terms + adding all the 3's in the ten's digit of the 9 terms (except first term which is just 3) + adding all the 3's in the hundred's digit of the 8 terms (except 3 and 33 which don't have a hundred's term)
= (3*10) + (30*9) + (300*8) = 2700 => hundred's terms is 7!
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