264. 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
a1,a2,a3...a15 In the sequence shown, a(n) = a(n-1)+k,
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Target question: How many term in the sequence a1,a2,a3...a15 are greater than 10?varun289 wrote:264. 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
Given: a(n) = a(n-1)+k, where 2<=n<=15
In other words, each term is derived by taking the term before it and adding k
IMPORTANT: Keep in mind that k can be either a positive or negative number. So, the sequence may be increasing (e.g., 5, 7, 9, 11...) or it may be decreasing (e.g., 20, 15, 10, ...)
Statement 1: a1 = 24
The 1st term is 24, but since we don't know the value of k, there's no way to determine the terms in the sequence that are greater than 10
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: a8 = 10
Let's consider the 2 possible cases for k (k is positive or k is negative)
case a: k is positive.
This means that the sequence is increasing.
In other words, term 1 < term 2 < term 3, etc.
The 8th term is 10, which means every term after the 8th term must be greater than 10.
So, terms 9, 10, 11, 12, 13, 14 and 15 are greater than 10.
This means that 7 terms in the sequence are greater than 10
case b: k is negative.
This means that the sequence is decreasing.
In other words, term 1 > term 2 > term 3, etc.
The 8th term is 10, which means every term before the 8th term must be greater than 10.
So, terms 1, 2, 3, 4, 5, 6 and 7 are greater than 10.
This means that 7 terms in the sequence are greater than 10
Since both cases yield the same answer to the target question, we can be certain that 7 terms in the sequence are greater than 10
Statement 2 is SUFFICIENT
Answer = B
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Brent
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I have a query here, If we take the possibility of k being positive and all other terms after a8 positive, still, we cannot say for certain that the earlier 7 are not positive, so we can have either 7 or all elements as positive.
Pls guide.
Thanks. Sonali
Pls guide.
Thanks. Sonali
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Hi Sonali,sonalibhangay wrote:I have a query here, If we take the possibility of k being positive and all other terms after a8 positive, still, we cannot say for certain that the earlier 7 are not positive, so we can have either 7 or all elements as positive.
Pls guide.
Thanks. Sonali
I believe that you're referring to the sufficiency of statement 2.
If k is positive, then we can be certain that the 9th term, the 10th term etc are all greater than 10. That's all that matters here.
What about the first 7 terms?
Well, they may all be positive, as in the case: 3,4,5,6,7,8,9,10, 11, 12, 13, 14, 15, 16, 17
Or some of them may be positive, as in the case: -11, -8, -5, -2, 1 , 4, 7, 10, 13, 16, 19, 22, 25, 28, 31
Or they may all be negative, as in the case: -130, -110, -90, -70, -50, -30, -10, 10, 30, 50, 70, 90, 110, 130, 150
But keep in mind that the target question has nothing to do with whether or not the first 7 terms are positive or not. The target question is asking how many terms are greater than 10.
Cheers,
Brent
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The sequence is an A.P. with common difference of k(a nonzero constant).
Statement 1) a1 = 24
We have no information about k being positive or negative and this is important as:-
If k > 0, then a2, a3, a4....a15 all greater than 24, hence > 10.
If k < 0, then depending upon the value of k only, we can figure out with certainty that how many terms are greater than 10.
Not Sufficient!!!
Statement 2) a8 = 10
Now a8 is the middle term and hence the median, so seven of the fifteen terms must be > 10 .
Sufficient!!!
Answer B.
Statement 1) a1 = 24
We have no information about k being positive or negative and this is important as:-
If k > 0, then a2, a3, a4....a15 all greater than 24, hence > 10.
If k < 0, then depending upon the value of k only, we can figure out with certainty that how many terms are greater than 10.
Not Sufficient!!!
Statement 2) a8 = 10
Now a8 is the middle term and hence the median, so seven of the fifteen terms must be > 10 .
Sufficient!!!
Answer B.
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if k is negative that means there are 6 terms NOT 7 that are greater than 10 correct? Because
2<= n =< 15
That means 1) n=2 2) n=3 3) n=4 4) n=5 5) n=6 6) n=7
And if k is positive than that means we have 7 terms that are greater than 10.
Hence the answer would be C right?
Please let me know if im missing something here.
Thanks!
2<= n =< 15
That means 1) n=2 2) n=3 3) n=4 4) n=5 5) n=6 6) n=7
And if k is positive than that means we have 7 terms that are greater than 10.
Hence the answer would be C right?
Please let me know if im missing something here.
Thanks!
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I'm not entirely sure what you mean by: 1) n=2 2) n=3 3) n=4 4) n=5 5) n=6 6) n=7jassim.arif wrote:if k is negative that means there are 6 terms NOT 7 that are greater than 10 correct? Because
2<= n =< 15
That means 1) n=2 2) n=3 3) n=4 4) n=5 5) n=6 6) n=7
And if k is positive than that means we have 7 terms that are greater than 10.
Hence the answer would be C right?
Please let me know if im missing something here.
Thanks!
Do you mean that, for term3, n=4? If so, this is not the case.
We're told that a(n) = a(n-1)+k
So, for term3, n = 3
For term5, n = 5. and so on.
Applying the ruke, we get ... term2 = term1 + k
term3 = term2 + k
term4 = term3 + k
etc.
Notice that, since the formula applies only to 2 < n < 15, it doesn't apply to the first term. This makes sense, because we'd get term1 = term0 + k and there is no term0 in this sequence.
Cheers,
Brent