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
Averages
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xcusemeplz2009
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uttam.albela
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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.
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.
-
Abdulla
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So simple !!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.
Abdulla
