find the average of first 71 natural numbers ?
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Subject:
Math -
Author:
dashawnmcmillan -
Created:
1 year ago
Answers 2
See AP (Arithmetic Progression)
a+(n-1)d=1+(71-1)1=71 [a=1,n=71,d=1]
Sum= =n[2a+(n-1)b]/2 =71[2+(71-1)1]/2=71[2+70]/2=(71×72)/2=5112/2=2556.
Average = sum/ no. of observations [here, no. of natural nos.]=2556/71=36.
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Author:
evaehqw
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Answer:
The answer is 36
Given problem:
find the average of first 71 natural numbers ?
Step-by-step explanation:
We know the first 71 natural number series is 1, 2, 3, 4... 71
Here 1, 2, 3, 4.. 71 is in the form of a AP
where first term [tex]a[/tex] = 1, last term [tex]l[/tex] = 71 and number of terms = 71
therefore, sum of terms = [tex]\frac{n(a+l)}{2}[/tex] = [tex]\frac{71(1+71)}{2}[/tex] = [tex]\frac{71(72)}{2}[/tex] = 71(36) = 2556
Average of first 71 natural numbers
= sum of 71 numbers / 71
= 2556 / 71 = 36
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Author:
whiz760z
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