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Question:     Find the average of even numbers from 10 to 1258


Correct Answer  634

Solution And Explanation

Solution

Method (1) to find the average of the even numbers from 10 to 1258

Shortcut Trick to find the average of the given continuous even numbers

The even numbers from 10 to 1258 are

10, 12, 14, . . . . 1258

After observing the above list of the even numbers from 10 to 1258 we find that the difference between two consecutive terms are equal. This means the list of the even numbers from 10 to 1258 form an Arithmetic Series.

In the Arithmetic Series of the even numbers from 10 to 1258

The First Term (a) = 10

The Common Difference (d) = 2

And the last term (ℓ) = 1258

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 1258

= 10 + 1258/2

= 1268/2 = 634

Thus, the average of the even numbers from 10 to 1258 = 634 Answer

Method (2) to find the average of the even numbers from 10 to 1258

Finding the average of given continuous even numbers after finding their sum

The even numbers from 10 to 1258 are

10, 12, 14, . . . . 1258

The even numbers from 10 to 1258 form an Arithmetic Series in which

The First Term (a) = 10

The Common Difference (d) = 2

And the last term (ℓ) = 1258

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 1258

1258 = 10 + (n – 1) × 2

⇒ 1258 = 10 + 2 n – 2

⇒ 1258 = 10 – 2 + 2 n

⇒ 1258 = 8 + 2 n

After transposing 8 to LHS

⇒ 1258 – 8 = 2 n

⇒ 1250 = 2 n

After rearranging the above expression

⇒ 2 n = 1250

After transposing 2 to RHS

⇒ n = 1250/2

⇒ n = 625

Thus, the number of terms of even numbers from 10 to 1258 = 625

This means 1258 is the 625th term.

Finding the sum of the given even numbers from 10 to 1258

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 1258

= 625/2 (10 + 1258)

= 625/2 × 1268

= 625 × 1268/2

= 792500/2 = 396250

Thus, the sum of all terms of the given even numbers from 10 to 1258 = 396250

And, the total number of terms = 625

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 1258

= 396250/625 = 634

Thus, the average of the given even numbers from 10 to 1258 = 634 Answer


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