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If the average(arithmetic mean) of 18 consecutive
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02 Oct 2018, 06:05
02 Oct 2018, 06:05
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72% (01:47) correct
27% (02:15) wrong based on 59 sessions
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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
A)517
B)518
C)519
D)521
E)525
Re: If the average(arithmetic mean) of 18 consecutive
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02 Oct 2018, 21:57
02 Oct 2018, 21:57
solution pls??
Re: If the average(arithmetic mean) of 18 consecutive
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03 Oct 2018, 04:19
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, \(x = 517\)
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, \(x = 517\)
