a(1), a(2), a(3), ……, a(15).

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

The answer is (C)

a(n) = a(n-1)+ k

from (1) a(2) = a(1) + k = 24 +k, but we don't what k is, So insufficient, similarly
from (2) a(8) = 10 = a(7) + k, but we don't know what a(7) and k is. So insufficient.

from question stem, we know: -

a(7) = a(6) + k
a(6) = a(5) + k and so on so a(7) = a(2) + 6k.......(3)

from (1) we know a(2) = a(1) + k = 24 +k
from (2) a(8) = a(7) + k so a(7) = 10-k

Putting the values of a(2) and a(7) in (3)
10-k = 24+6k

so k = -2, now no of terms greater than 10 can be calulated.

So (C) is the correct answer

The answer is actually (B) and of course I couldn't think about this when I took the test ;-)

If a(n) = a(n-1) + k, then a(1), a(2),.....a(15) is an arithmetic progression.

Statement (2) says a(8) = 10

a(8) is the middle term and hence the Median of this series. The number k can be positive or negative. If k is negative, then the numbers a(1), a(2) etc. will be in descending order and if it is positive, then the series will be in ascending order. No matter what, there will be 7 terms which will be greater than 10 (Please note that since k is non-zero, and hence all the terms in the series will be DISTINCT).

It could a(1) to a(7) if k is negative, and a(9) to a(15) if k is positive.

This is definitely a trap kind of questions.

That's right Harish. Infact, while answering the question, it almost caught my eye, but I was in flow so didn't realize. Thanks for the posting good questions. This one certainly has a trap. :)

GMAT 2007

How is a(8) the middle term? Isn't the middle term of this set the average of a(8) and a(9)?

nope.

most convincing argument: go ahead and list them yourself, and count. there are seven terms before a8, and seven terms after it. therefore, it's the middle term.
if you draw a line between a8 and a9, there will be eight terms to the left of the line and seven terms to the right of the line. therefore, that line is not the middle of the sequence.

--

general facts: (let n stand for the number of terms)

* if you have an even # of terms
in this case, the 'middle term' (median) is the average of the two middle terms, which are term number (n/2) and term number (n/2 + 1).

* if you have an odd # of terms
in this case, the middle term is just one term: term number (n + 1)/2.

hope that helps