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Re: If the average(arithmetic mean) of 18 consecutive
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03 Oct 2018, 04:19
2
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, \(x = 517\)
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, \(x = 517\)
Re: If the average(arithmetic mean) of 18 consecutive
[#permalink]
18 Nov 2018, 05:51
1
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, \(x = 517\)
Re: If the average(arithmetic mean) of 18 consecutive
[#permalink]
10 Jan 2019, 16:45
2
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
Re: If the average(arithmetic mean) of 18 consecutive
[#permalink]
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, \(x = 517\)
Re: If the average(arithmetic mean) of 18 consecutive
[#permalink]
04 Sep 2024, 10:34
Expert Reply
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: