Sequence A is defined by the equation An = 3n + 2, where n is an integ

What if i take min = 11 and as range = 30 implies B = 41
so set = {11,14,17,20,23,26,29,32,35,38,41}
than the median = 26

what i am confused here is that in such kind of Q, are they asking if we can find the median based on the given information or not.
Or
are we suppose to determine that with given info there exists only single median (you know what i mean)?


But now when i see statement 1, according to which also we can construct different sets, just that the sum of all elements should be 220. So basically i think the Q is asking if we can find the median based on the given info or not.

Based on the given info, you can just create ONE set:
set B can't start from 11.
We know
A={5,8,11,14....}
AND it is given in the stem that:
"If set B is comprised of the first x terms of sequence A".
Thus, B can't start from 11.

And now, when a series is starting with 5, we know this for sure. We also know the summation of the series from st1. We will know the exact set to form. Once we know the set, we can find the median.

Did I answer your question?

one thing we must remember is in DS the 2 statements can never contradict each other.

If your statement 2 stats with 5..................41
then the st.1 should mention the sum of set B as 226 and not 220.

Hope that helps.

Sequence is like 5 8 .....

1. Sufficient

sum of terms is 220

so enough to find the last term and all the terms upto the last term (as the series is in A.P.)

=> enough to find median

2. Sufficient

from the range and the first term , we can find the last term and all the terms upto the last term (as the series is in A.P.)

=> enough to find median

Answer is D.

a1=5,a2=8,a3=11 and so on.

a gives 120 = n/2[2a+ (n-1)d] where d = 3 and a = 5 here.

thus the xth element can be found out. Hence the median.

Sufficient.

b gives a = 5 and xth element = 35.
thus median can be found out for AM.
Sufficient.

D it is.

Hi Bunnel and Fluke,

The problem reads that Sequence A is defined by the equation An = 3n + 2, therefore Sequence A is definitely in Arithmetic progression.

Then it reads If set B is comprised of the first x terms of sequence A- How can we assume from this that set B in Arithmetic progression.

As per Fluke's solution, statement A is only sufficient if it is assumed that set B is in AP.

From statement B we know that the t1= 5 and tn=35

set B could be = {5, t2, t3, t4, ...t(n-1),35}

But if we combine both the information then

Assuming that set B is not in AP

5+t2+t3+t4+...+t(n-1)+35=220
t2+t3+t4+...+t(n-1)=180

Therefore set B could be= (5,30,30,30,30,30,30,35) [as the problem does not read that set B consists of different integers], median 30
or set B could be=(5,24,28,29,31,32,36), median 30
or set B could be= (5,7,16,18,20,22,27,30,35), median 20

The way I understand the problem, if we assume that set B is in AP then the answer choice is D
If we assume that set B is not in AP and can consist of both different or same integers then the answer choice is E

It would be very helpful if you can please provide some insight on my understanding.

Set A = {5, 8, 11, 14, 17, 20, ...}, an arithmetic progression as you noted.

Set B is comprised of the first x terms of set A, so set B must also be an arithmetic progression. For example, if x=3, so if set B is comprised of the first 3 terms of set A, then set B = {5, 8, 11}.

Does this make sense?

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