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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

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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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

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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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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blackjack: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Tue May 04, 2010 6:12 am; Thanked: 2 times

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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kitg: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Mon May 02, 2011 1:32 pmFollowed by: 2 members; Location: Bangalore; GMAT Score:700

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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