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by pradeepkaushal9518 » Tue May 04, 2010 7:19 am
if the average (arithmetic mean) of 18 consecutive odd integers is 534,then the least of these integers is

1. 517
2.518
3.519
4.521
5.525

plz explain yr answers
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by rn4gmat » Tue May 04, 2010 8:15 am
Let the first term be X
The 18th term will be : X+34

So :

X + X + 34 = 2*534

2X = 1068-34 = 1034

so X = 517.

So the least term is 517.

Hope this helps.
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by kstv » Tue May 04, 2010 8:20 am
Least no = x
18th no = x +34
x+x+34 = (534)*2 = 1068
2x = 1034
x = 517
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by pradeepkaushal9518 » Tue May 04, 2010 8:23 am
thanks buddy
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by akhpad » Tue May 04, 2010 8:23 am
Let first term be X
X, X+2, X+4, X+6
Sum = 18X + (2+4+6+ ... 17th terms) = 18X + 2(1+2+3 ... 17th terms) = 18X + 17*18
Average = X + 17 = 534
X = 517

Answer: A
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by sk818020 » Tue May 04, 2010 8:32 am
If you don't want to do any math you can solve this one just by realizing that the average of any even set of odd integers will also be the median and that the the 9th and 10th members of that set will be 1 less and 1 more than the average. For this, the 9th = 533 and 10th = 535.

Counting down to the 1st element, 531=8th, 529=7th, 527=6th, 525=5th, 523=4th, 521=3rd, 519=2nd, 517=1st.
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by pradeepkaushal9518 » Tue May 04, 2010 8:35 am
f you don't want to do any math you can solve this on just by realizing that the average of any even set of odd integers will also be the median and that the the 9th and 10th members of that set will be 1 less and 1 more than the average. For this, the 9th = 533 and 10th = 535.

Counting down to the 1st element, 531=8th, 529=7th, 527=6th, 525=5th, 523=4th, 521=3rd, 519=2nd, 517=1st.

ya buddy for yr method
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