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If the average(arithmetic mean) of 18 consecutive

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If the average(arithmetic mean) of 18 consecutive [#permalink] New post 02 Oct 2018, 06:05
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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
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post 02 Oct 2018, 21:57
solution pls??
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post 03 Oct 2018, 04:19
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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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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post 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 ?
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post 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
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post 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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