Given a sequence: a1, a2, a3, ... a20, a21

Question

Given a sequence:  a_1 ,  a_2 ,  a_3 , ...  a_{20} ,  a_{21} 
In the sequence shown,  a_n = a_{n-1} +k, where  2\leq{n}\leq{21}  and k is a nonzero constant. How many of the terms in the sequence are greater than 15?

(1)   a_1 =35
(2)  a_{11} =15

nikchau · Answer

Usually, the answer to these questions would be C), since you would require both the initial term and one other term in the sequence to calculate the common difference i.e. k in this case.

However, in this question the middle term of the sequence itself is 15, and the question asks how many terms are greater than 15, the answer would be 10 terms:
i.e) a1-10 in case of a -ve value for k, or
i.e) a12-a21 in case of a +ve value for k

Therefore statement 2 is enough and ther correct answer would be B)

Mantrix · Answer

here in this question
a1 = 35
and the a2 would be a1 + k, a3 would be a2+k

(By A only we won't be able to conclude anything about how many terms are greater than 15), so reject the A and D option

Now a11 = 15 is given in the question, 
here we are observing that K is negative constant and it would be going to decrease the value of the expression as we process, and once we reach to 0, that will also lead us to the negative side of the number line

that means only till a10 we have numbers greater than 15

hence Answer is B

SatvikVedala · Answer

So from B if k is positive a12 to a21 are greater than 15
if k is negative a1 to a11 are lesser than 15
So Statement B alone is sufficient

pierjoejoe · Answer

the series is such that k is non zero.
statement 1 is not helpful, k might be any values and thus #of values greater than 15 might be anything. NOT SUFFICIENT
statement 2 is telling us that the median of the series (a11) is actually 15. there are 2 cases
 case 1 
-> all the values are the same --> all the values are 15 --> #values greater than 15 = all the values--> but k in this case is equal to 0 --> not possible per problem statement (k is non zero)
 case 2 
-> k is different from 0, whatever is the values of k (negative or positive), a11 is still the median, thus the middle values--> #values greater than 15 = 10 always by the definition of median. SUFFICIENT

bumpbot · Answer

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Given a sequence: a1, a2, a3, ... a20, a21

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