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tough Q

Expert replies
by atulmangal » Fri May 06, 2011 7:02 am
In the sequence shown above, a(n) = a(n-1) + k, where 2 <= n <= 15 and k is a non-zero constant. How many terms in the sequence are greater than 10?

(1) a(1) = 24 (2) a(8) = 10
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Source: — Data Sufficiency |

by MAAJ » Fri May 06, 2011 7:37 am
Edit: Oh never mind... It's a DS question lol... I'll go for...C???


Is OA 6??? the question doesn't say that n must be an integer but I'm assuming that...
Because when n = 8 the result is 10, all the values less than n = 8 will yield a result greater than 10. This is n = 2,3,4,5,6,and 7.

I did the math, but I think it's useless in this question

Recursive formula:
a(1) = a(0) + k = 24
a(8) = a(7) + k = 10

a(8) = a(1) + 7k = 10
a(8) = 24 + 7k = 10
7k = 10 - 24
7k = -14
k = -2

Lineal formula:
a(n) = kn + x
a(n) = -2n + x
a(1) = -2 + x = 24
-2 +x = 24
x = 24+2 = 26

Final formula -> a(n) = -2n + 26

atulmangal wrote:In the sequence shown above, a(n) = a(n-1) + k, where 2 <= n <= 15 and k is a non-zero constant. How many terms in the sequence are greater than 10?

(1) a(1) = 24 (2) a(8) = 10
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
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by ankurmit » Fri May 06, 2011 8:45 am
atulmangal wrote:In the sequence shown above, a(n) = a(n-1) + k, where 2 <= n <= 15 and k is a non-zero constant. How many terms in the sequence are greater than 10?

(1) a(1) = 24 (2) a(8) = 10
I could not understand your quistion..
--------
Ankur mittal
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by clock60 » Fri May 06, 2011 11:45 am
hi guys
it seems to me that i saw this prolem ealier so i `ii try to say why answer is B
clearly 1 st does not say anything, to the 2 st
(2) a(8)=10
a(n)=a(n-1)+k
a(8)=10
a(9)=a(8)+k, if k>0 then
a(9)=10+k
a(10)=10+2k.... and so on
so it happens that if k>0,each term after a(8) will be greater then 10, and in total
a9.a10.a11.a12.a13.a14-6 terms in total
if k<0 then,again
a(9)=a(8)+(-k)=10-k
a(10)=10-2k
a(11)=10-2k... and so on if k<0 then each term after a(8) will be less then 10, but each term before a(8) will be greater then 10, to prove
a(8)=a(7)+(-k). 10=a(7)-k. a(7)=10+k, and in similar way
a(6)=10+2k
a(5)=10+3k....and so on, as 2<=n<=14
a7,a6,a5,a4,a3,a2 will be greater than 10, notice that we do not include a1 an n>=2
and again in total we got six terms
so B is suff
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by atulmangal » Fri May 06, 2011 2:39 pm
Hey clock...

Brilliant Explanation....OA is indeed B

What do u think about the difficulty level of this question..????
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by clock60 » Fri May 06, 2011 10:45 pm
hi Atul
thanks for kind words
as about difficulty, to me it is pretty hard and even unsolvable if you see this for the first time
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by atulmangal » Fri May 06, 2011 10:59 pm
clock60 wrote:hi Atul
thanks for kind words
as about difficulty, to me it is pretty hard and even unsolvable if you see this for the first time
Hi yeah i say this question is tricky...i picked up Op C first time because using both we can calculate the exact value of K, and i thought without getting exact value of K we can't say anything but we don't need the value of K...its a kind of C trap question when both Op's together seems an obvious answer...
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