Question:
Find the average of even numbers from 10 to 1818
Correct Answer
914
Solution And Explanation
Solution
Method (1) to find the average of the even numbers from 10 to 1818
Shortcut Trick to find the average of the given continuous even numbers
The even numbers from 10 to 1818 are
10, 12, 14, . . . . 1818
After observing the above list of the even numbers from 10 to 1818 we find that the difference between two consecutive terms are equal. This means the list of the even numbers from 10 to 1818 form an Arithmetic Series.
In the Arithmetic Series of the even numbers from 10 to 1818
The First Term (a) = 10
The Common Difference (d) = 2
And the last term (ℓ) = 1818
The average of the numbers forming an Arithmetic Series
= The first term (a) + The last term (ℓ)/2
⇒ The average of numbers forming an Arithmetic Series = a + ℓ/2
Thus, the average of the even numbers from 10 to 1818
= 10 + 1818/2
= 1828/2 = 914
Thus, the average of the even numbers from 10 to 1818 = 914 Answer
Method (2) to find the average of the even numbers from 10 to 1818
Finding the average of given continuous even numbers after finding their sum
The even numbers from 10 to 1818 are
10, 12, 14, . . . . 1818
The even numbers from 10 to 1818 form an Arithmetic Series in which
The First Term (a) = 10
The Common Difference (d) = 2
And the last term (ℓ) = 1818
The Average of the given numbers
= Sum of the given numbers/Total number of given numbers
Thus, to find the average of the given numbers, first, we need to find their sum and the total number of given numbers
Finding the number of terms
For an Arithmetic Series, the nth term
an = a + (n – 1) d
Where
a = First term
d = Common difference
n = number of terms
an = nth term
Thus, for the given series of the even numbers from 10 to 1818
1818 = 10 + (n – 1) × 2
⇒ 1818 = 10 + 2 n – 2
⇒ 1818 = 10 – 2 + 2 n
⇒ 1818 = 8 + 2 n
After transposing 8 to LHS
⇒ 1818 – 8 = 2 n
⇒ 1810 = 2 n
After rearranging the above expression
⇒ 2 n = 1810
After transposing 2 to RHS
⇒ n = 1810/2
⇒ n = 905
Thus, the number of terms of even numbers from 10 to 1818 = 905
This means 1818 is the 905th term.
Finding the sum of the given even numbers from 10 to 1818
The sum of all terms (S) in an Arithmetic Series
= n/2 (a + ℓ)
Where, n = number of terms
a = First term
And, ℓ = Last term
Thus, the sum of all terms (S) of the given even numbers from 10 to 1818
= 905/2 (10 + 1818)
= 905/2 × 1828
= 905 × 1828/2
= 1654340/2 = 827170
Thus, the sum of all terms of the given even numbers from 10 to 1818 = 827170
And, the total number of terms = 905
Since, the average of the given numbers
= Sum of the given numbers/Total number of given numbers
Thus, the average of the given even numbers from 10 to 1818
= 827170/905 = 914
Thus, the average of the given even numbers from 10 to 1818 = 914 Answer
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