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

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Joined: 04 Jan 2015
Posts: 2438
The median of n consecutive odd integers is 30. If the fifth term.....  [#permalink]

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Updated on: 29 Oct 2018, 22:38
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95% (hard)

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21% (01:50) correct 79% (01:53) wrong based on 120 sessions

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The median of n consecutive odd integers is 30. If the fifth term is 33, then which of the following can be the last term of the sequence?

A. 19
B. 21
C. 25
D. 33
E. 41

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Originally posted by EgmatQuantExpert on 25 Oct 2018, 00:22.
Last edited by EgmatQuantExpert on 29 Oct 2018, 22:38, edited 1 time in total.
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Posts: 7200
Re: The median of n consecutive odd integers is 30. If the fifth term.....  [#permalink]

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25 Oct 2018, 04:23
The median of n consecutive odd integers is 30. If the fifth term is 33, then which of the following can be the last term of the sequence?

A. 19
B. 21
C. 25
D. 33
E. 41

Now median = 30, so two middle terms are 29 to 31
Fifth term is 33..
1) if the sequence is increasing...
33 as fifth term means ..... 25,27,29,31,33
So 3 terms either side of median, last term becomes 35, but that is not in the choices
2) if the sequence is decreasing..
33 is fifth term , so 31 is sixth term and therefore 6 terms on either side of median 30..
Total terms - 12, so last term = 33-(12-5)*2=33-14=19
So sequence 41,39,37,35,33,31,29,27,25,23,21,19

A
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Re: The median of n consecutive odd integers is 30. If the fifth term.....  [#permalink]

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25 Oct 2018, 22:24
I have one query here.
Median is calculated when we arrange all elements in increasing order.
Am I wrong?
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Posts: 2438
Re: The median of n consecutive odd integers is 30. If the fifth term.....  [#permalink]

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29 Oct 2018, 00:39

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

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Joined: 12 Sep 2017
Posts: 3
Re: The median of n consecutive odd integers is 30. If the fifth term.....  [#permalink]

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29 Oct 2018, 04:24
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
Re: The median of n consecutive odd integers is 30. If the fifth term..... &nbs [#permalink] 29 Oct 2018, 04:24
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# The median of n consecutive odd integers is 30. If the fifth term.....

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