an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant

How many of the terms in the sequence are greater than 10?

Statement 1: a1 = 24

Since we dont know k, we cannot know how many terms will be greater than 10.

for eg.
-if k > 0, then all 15 terms will be greater than 10.
-if k = -20, then only the first term, a1, is greater than 10.

Insufficient.

Statement 2: a8 = 10

Notice that a8 is the middle term of the 15-term sequence. So this means there are 7 terms greater than 10, and 7 terms smaller than 10 (notice that this true regardless of what k is, as long as k is nonzero). We have sufficient info to answer this question.

Choose B.

-BM-

I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?

aj5105 wrote:I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?

1. It depends whether the common difference (in this case case K is common differnce) is positive or negative

2. As K is non-zero therefore, either the terms will be in ascending (if k is +ve) or descending (if k is -ve)

aj5105 wrote:I have a couple of questions here. They may be stupid, but please bear with them.

1.Are the terms in a sequence always arranged in ascending order?

2.Is there a possibility that all terms from a8 to a15 remain same?

aj5105,
I agree with bluementor about the answer B
as for your questions:
1) No, a sequence is not always in an order. It depends upon the sequence's formula or function. Arithmetic or geometric progression are just examples of sequence

2) It is possible but not in this case. it is clearly mentioned that K is a non-zero constant. if K can be zero then of course all the terms are equal

is there a typo in this question? how can there be 15 sequences if 2 ≤ n ≤ 15 ??

aj5105 wrote:Given a sequence a1, a2, a3��. a15

In the sequence shown, an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant. How many of the terms in the sequence are greater than 10?

(1) a1 = 24
(2) a8 = 10

Tricky question..almost got me..worth solving...
[spoiler]
takeaway : C-Trap says " In DS, Always be suspicious of answers that you can get using 2 options without using your pen i.e. too obvious looking answer." - This takeaway saved me in this question.[/spoiler]

Given a sequence a1, a2, a3??. a15

In the sequence shown, an = an-1 + k, where 2 =< n =<15 and k is a non-zero constant. How many of the terms in the sequence are greater than 10?

(1) a1 = 24

a1= 24
a2=24+k
a3=24+2k
.....
We cannot say how many are greater than 10, because, we do not know if K is positive or negative.

(2) a8 = 10
Series is linear.. Whether it needs to increase or decrease linearly.
a1,a2,..........a8,..............a14,a15
On either side of a8, there are only 7 terms.
So either a1-a7 or a9-a15 need to be less than 10.

hence pick B.

I was almost getting it wrong when I looked at why 2<=n<=15 is given. That made ot easy for me.

we dont have 7 terms on either side of a8.Thereare 6 terms on the left of a8 and 7 to the right
2,3,4,5,6,7
9,10,11,12,13,14,15

clammiestqasar wrote:we dont have 7 terms on either side of a8.Thereare 6 terms on the left of a8 and 7 to the right
2,3,4,5,6,7
9,10,11,12,13,14,15

The series is given as a1,a2,a3...

yea..apologies for being so stupid.