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Averages

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by Abdulla » Wed Oct 21, 2009 5:19 pm
There is a set of 160 numbers, beginning at 6, with each subsequent term increasing by increment of 3. what is the average of this set of numbers?

OA is 244.5
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by xcusemeplz2009 » Wed Oct 21, 2009 8:40 pm
the sequence is 6,9,12,18.............

last term i.e a160=a1+d(n-);a1=6;d=3;n=160
a160=477

sum of the series is n(a1+a160)/2=160*(6+477)/2=489*80

avg=sum/160 i.e 489*80/160=244.5
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by uttam.albela » Wed Oct 21, 2009 9:01 pm
Hi Abdulla,

Whenever you have a series of numbers with fixed difference between any two consecutive terms, the average is AVG of FIRST and LAST term.

first term = 6

last term = 6 + 3 * (160-1)= 6 + 3*159 = 483

Avg = (6 + 483)/2 = 489 / 2 = 244.5

Any doubt in it, you r most welcome to discuss.
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by Abdulla » Thu Oct 22, 2009 6:48 pm
uttam.albela wrote:Hi Abdulla,

Whenever you have a series of numbers with fixed difference between any two consecutive terms, the average is AVG of FIRST and LAST term.

first term = 6

last term = 6 + 3 * (160-1)= 6 + 3*159 = 483

Avg = (6 + 483)/2 = 489 / 2 = 244.5

Any doubt in it, you r most welcome to discuss.
So simple !! :D
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by NikolayZ » Sat Oct 24, 2009 12:12 pm
I started solving this one finding 79th and 80th terms.
Then we have to add them up and divide by 2 to find the average!
79th=6+79*3=237+6=243
80th=6+80*4=246
so average equals (243+246)/2 =244,5
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