The median of n consecutive odd integers is 30. If the fifth term.....

Solution

Given:

• The median of n consecutive odd integers = 30
• The fifth term = 33

To find:

• The last term of the sequence

Approach and Working:

• Let us assume that the ‘n’ consecutive odd integers are {a, a + d, a + 2d, a + 3d, ……a + (n-1)d}
• The ‘n’ consecutive odd integers can be arranged in either decreasing order or increasing order
• Per our conceptual knowledge, the difference between any two consecutive odd integers is 2

Case 1: They are arranged in increasing order

• In this case, the value of d = 2, since they are arranged in increasing order
• Given that the median of this set = 30, which is an even number

o For median to be an even number, the number of terms must be even,
o That is, median = average of middle two terms = \(\frac{(odd + odd)}{2}\)

Note: If the number of terms is odd, then median = the middle term, which is odd

• Thus, median = \((t_{n/2} + t_{n/2 + 1})/2 = 30\)

o Implies, \(t_{n/2} = 29\) and \(t_{n/2 + 1} = 31\)
o \(t_{n/2} = a + (\frac{n}{2} -1)d = a + \frac{nd}{2} – d = 29\)

• Substituting d = 2 in the above equation, we get, a + n – 2 = 29

o Implies, a + n = 31 ……………………………………………… (1)

• We are also given that \(t_5 = 33\)

o Thus, a + (5 - 1)*2 = 33
o Therefore, we get the value of a = 33 – 8 = 25

• Substituting this value of a in equation (1), we get, n = 31 – 25 = 6
• Therefore, the last term = a + (6 - 1)*d = 25 + 10 = 35

But we do not have any such answer choice, so let’s see the other case.

Case 2: They are arranged in decreasing order

• In this case, the value of d = -2, since they are arranged in decreasing order
• Thus, median = \((t_{\frac{n}{2}} + t_{\frac{n}{2} + 1})/2 = 30\)
• Implies, \(t_{\frac{n}{2}} = 31\) and \(t_{\frac{n}{2} + 1} = 29\)
• \(t_{\frac{n}{2}} = a + (\frac{n}{2} -1)d = a + \frac{nd}{2} – d = 31\)

o Substituting the value of d as -2, we get, a - n + 2 = 31

 Implies, a - n = 29 ……………………………………………… (2)

• We are also given that \(t_5 = 33\)

o Which implies, a + (5 -1) * (-2) = 33
o Therefore, a = 33 + 8 = 41

• Substituting this value of a in equation (2), we get, n = 41 – 29= 12
• Therefore, the last term = a + (12 - 1)*d = 41 - 22 = 19

Hence, the correct answer is option A.

Answer: A

As given ,

Median = 30

Since consecutive numbers , they are in AP
So ,

first term (a)+ last term (l) / 2 = 30
therefore ,

a+l = 60 -----------(i)

now ,

fifth term = a +(n-1)d
33= a+ 4d
a= 33 -4d------------(ii)

Substituting value of a in equation i

33-4d+l=60
l= 27 - 4d

Since it is odd number we can take common difference as ( 1,2,3,5,7 ) and check answer

therefore ,

l = 27- 4X2
l= 27-8
l= 19

Simple as that. Great Question

yes, you have to arranage in ascending or descending

But when the question says consecutive odd integers, it is implied that the sequence is increasing. If the sequence is decreasing, the question must make it clear that it is a the elements in the sequence are decreasing consecutively.

I assume you mean a consecutive integer sequence in the problem. Consecutive integers 'MUST' be an increasing sequence unless mentioned otherwise. The GMAT version is as below:

"The numbers - 2, - 1, 0, 1, 2, 3, 4, 5 are consecutive integers. Consecutive integers can be represented by n,n+1,n+2,n+3,...,where n is an integer.The numbers 0,2,4,6,8 are consecutive even integers, and 1, 3 , 5 , 7 , 9 are consecutive integers . Consecutive even integers can be represented by 2 n , 2 n + 2 , 2 n + 4 , . . . , and consecutive odd integers can be represented by 2n + 1, 2n + 3, 2n + 5, .. ., where n is an integer. "

Going by this stated version along with the lack of any qualified statement on the sequence in the problem statement, the sequence has to be increasing in nature.

Now, if my assumption is wrong and you don't mean a sequence of consecutive integers, the other possibility could be that they are just a set of consecutive integers arranged in a pattern. Since the pattern is not mentioned, there can be umpteen number of possible sequencing even while satisfying the other given conditions in the problem. For example:
The sequence could be defined by a very crude function as:

A(n) = 23, if n=1;
A(n) = 27, if n=2;
A(n) = 21, if n=3;

So, the last term ('not the min or max') could be any odd number between 21 and 43 (with only a few exclusions).

Please do let me know if I am horribly wrong anywhere in my reasoning.

how can consecutive odd number contains 30 as median because 30 is even number i didn't understand that ?­

The median of an even count of elements is the average of the two middle terms when they are in order. For example, the median of {29, 31} is (29 + 31)/2 = 30.