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Terms in a series greater than 10

Expert replies
by vivekjaiswal » Fri Nov 27, 2009 8:29 pm
a1,a2,a3...a15
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA is B

I don't get this at all...please help.
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Source: — Data Sufficiency |

by Testluv » Sat Nov 28, 2009 1:23 am
vivekjaiswal wrote:a1,a2,a3...a15
In the sequence shown, a(n) = a(n-1)+k, where 2<=n<=15 and k is a nonzero constant. How many of the terms in the sequence are greater than 10?
(1) a1 = 24
(2) a8 = 10

OA is B

I don't get this at all...please help.
So, we know from the question stem that it is an arithmetic sequence. You get each term by adding k to the previous term. For example, the sequence could be: 1, 4, 7, 10, and in this case k = 3.

(1) a1 = 24

The first term is 24. If you thought this meant that all the terms were greater than 10, then you assumed that k was positive. Remember that all we know from the question is that k is a nonzero constant. So, k, in fact, can be negative. If k were negative we would have a "shirking" or decreasing sequence instead of an increasing one.

Because we don't know k's sign, the first statement is insufficient.

(2) a8 = 10

In this case, if k is positive (increasing sequence), then only the the last seven terms (ie, a9 to a15) are all greater than 10.

But, if k is negative (decreasing sequence), then only the first seven terms (ie, a1 to a7) are all greater than 10.

In both cases, the number of terms greater than 10 is seven (to answer the question, it doesn't matter which terms, it matters how many).

The second statement is independently sufficient while the first one is not.

Choose B.
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by vivekjaiswal » Sat Nov 28, 2009 2:08 am
Thanks Testluv for the explanation.
I was counting the terms from a2, as 2<=n<=15 is given. But now I realise that the condition is for the expression a(n)=a(n-1)+k.

Great explanation, and now I understand why it is important not to work even the simplest problems in your head :)

Thanks,
Vivek.
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by mohank80 » Wed May 25, 2011 4:32 pm
Here is one thing I notice. The values of n : 2<=n<=15.

Would this mean if we assume K is negative, n would be 2, 3, 4, 5, 6 and 7 (6 values)?
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