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# If the average(arithmetic mean) of 18 consecutive

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Director
Joined: 07 Jan 2018
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If the average(arithmetic mean) of 18 consecutive [#permalink]  02 Oct 2018, 06:05
1
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Question Stats:

70% (01:29) correct 29% (02:14) wrong based on 27 sessions
If the average(arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

A)517
B)518
C)519
D)521
E)525
[Reveal] Spoiler: OA
Intern
Joined: 07 Sep 2018
Posts: 29
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Kudos [?]: 2 [0], given: 5

Re: If the average(arithmetic mean) of 18 consecutive [#permalink]  02 Oct 2018, 21:57
solution pls??
Director
Joined: 07 Jan 2018
Posts: 648
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Kudos [?]: 606 [2] , given: 88

Re: If the average(arithmetic mean) of 18 consecutive [#permalink]  03 Oct 2018, 04:19
2
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There are altogether $$18$$ numbers.
Let us assume the first number is $$x$$ this leaves us with 17 numbers.
Since the numbers are consecutively odd such as 3,5,7 or 11,13,15.
Notice that for odd numbers there is a gap of 2 between successive numbers hence each number after the first will be 2 more than the previous number.
Hence the numbers will be such as x,x+2,x+2+2,x+2+2+2....
therefore if the first number is x the last number or the 18th number will be $$17*2 + x$$
since each number in the sequence are equally spaced the mean will be the avg of the first and last number.
which is $$\frac{x+x+38}{2} = \frac{2(x+17)}{2} = [m]x+17$$[/m]
From question $$x+17 = 534$$
therefore, $$x = 517$$
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Director
Joined: 09 Nov 2018
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Kudos [?]: 27 [0], given: 1

Re: If the average(arithmetic mean) of 18 consecutive [#permalink]  17 Nov 2018, 20:11
amorphous wrote:
There are altogether $$18$$ numbers.
Let us assume the first number is $$x$$ this leaves us with 17 numbers.
Since the numbers are consecutively odd such as 3,5,7 or 11,13,15.
Notice that for odd numbers there is a gap of 2 between successive numbers hence each number after the first will be 2 more than the previous number.
Hence the numbers will be such as x,x+2,x+2+2,x+2+2+2....
therefore if the first number is x the last number or the 18th number will be $$17*2 + x$$
since each number in the sequence are equally spaced the mean will be the avg of the first and last number.
which is $$\frac{x+x+38}{2} = \frac{2(x+17)}{2} = [m]x+17$$[/m]
From question $$x+17 = 534$$
therefore, $$x = 517$$

Is it (x+x+38)/2 or (x+x+34)/2 ?
Director
Joined: 07 Jan 2018
Posts: 648
Followers: 7

Kudos [?]: 606 [0], given: 88

Re: If the average(arithmetic mean) of 18 consecutive [#permalink]  18 Nov 2018, 05:51
AE wrote:
amorphous wrote:
There are altogether $$18$$ numbers.
Let us assume the first number is $$x$$ this leaves us with 17 numbers.
Since the numbers are consecutively odd such as 3,5,7 or 11,13,15.
Notice that for odd numbers there is a gap of 2 between successive numbers hence each number after the first will be 2 more than the previous number.
Hence the numbers will be such as x,x+2,x+2+2,x+2+2+2....
therefore if the first number is x the last number or the 18th number will be $$17*2 + x$$
since each number in the sequence are equally spaced the mean will be the avg of the first and last number.
which is $$\frac{x+x+38}{2} = \frac{2(x+17)}{2} = [m]x+17$$[/m]
From question $$x+17 = 534$$
therefore, $$x = 517$$

Is it (x+x+38)/2 or (x+x+34)/2 ?

Thanks for checking. It is (x+x+34)/2
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Director
Joined: 09 Nov 2018
Posts: 508
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Kudos [?]: 27 [1] , given: 1

Re: If the average(arithmetic mean) of 18 consecutive [#permalink]  10 Jan 2019, 16:45
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n be the first term
Last term => $$n+17*2=> n+34$$

Now as the series will be in Arithmetic progression
Mean = Average of the first and the last term
Hence 534 =$$\frac{n+n+34}{2}=> n+17$$
Hence n+17=534=> n=517

Hence A
Re: If the average(arithmetic mean) of 18 consecutive   [#permalink] 10 Jan 2019, 16:45
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