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If the average(arithmetic mean) of 18 consecutive

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amorphous
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If the average(arithmetic mean) of 18 consecutive [#permalink] New post   02 Oct 2018, 06:05
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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
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A
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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
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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
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If the average(arithmetic mean) of 18 consecutive [#permalink] New post   16 Sep 2022, 23:03
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Given that the average(arithmetic mean) of 18 consecutive odd integers is 534 and we need to find the least of these integers

============================================================

Theory
    ‣‣‣ In Case of consecutive number with even number of term, Mean = Mean of Middle two terms

============================================================

Let the middle two terms are 2x-1 and 2x+1

=> Mean = mean of middle two terms = \(\frac{2x-1 + 2x+1}{2}\)= \(\frac{4x}{2}\) = 2x = 534

As there are 18 terms so the middle two terms will be \(9^{th}\) and \(10^{th}\) term
=> \(9^{th}\) term = 2x-1 = 534-1 = 533

First term or the lest term will be \(9^{th}\) term - 2*8 = 533 - 16 = 517

So, Answer will be A.
Hope it helps!

Watch the following video to MASTER Statistics

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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post   04 Sep 2024, 05:23
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quickest way is that - you can see if its 18 ODD consecutive integers so if 534 is the average
then- the 9th and 10th integers are 533 and 535

now simply count backwards to 517
therefore A
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post   04 Sep 2024, 10:31
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, \(x = 517\)



Didnt Get it..Can you Please Explain Bit Moree?
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post   04 Sep 2024, 10:34
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Since the average of 18 consecutive odd integers is 534, the average of the middle two integers is also 534. Therefore, the middle two integers are 533 and 535. There will be 8 odd integers less than 533 (and 8 more greater than 535). So the smallest of these integers is 533 - 8(2) = 533 - 16 = 517.

Alternate Solution:

If the smallest of these integers is x, then the largest of these integers will be x + 34. Since consecutive odd integers form an evenly spaced set of integers, the average is equal to the average of the smallest and the largest integers; thus:

(x + x + 34)/2 = 534

2x + 32 = 1068

2x = 1034

x = 517

Answer: A
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post   04 Sep 2024, 10:34
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Let n be the first term
Last term => \(n+17*2=> n+34\)

Now as the series will be in AP.
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
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink] New post   19 Dec 2024, 21:04
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an easier method is to realize that we can subtract 500 from the average to get 34. This will mean that every singly number is subtracted by 500 as well. 34 is an even number so therefore, it is the average of the 9th and 10th terms. since all terms are evenly spaced, we can say that the 9th and 10th terms are 33 and 35 respectively. Now, let's work recrusively to get the first 9 terms
17,19,21,23,25,27,29,31,33
the least of the values in this list is 17, add 500 to this to match the ans choices to get 517

A
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Re: If the average(arithmetic mean) of 18 consecutive [#permalink]
19 Dec 2024, 21:04
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