Question

The sum of 38 consecutive positive whole numbers is 2,071. What is the first number?

Answer 1

Let the first number in the sequence of 38 consecutive positive whole numbers be denoted as \( x \). The numbers can then be represented as:

\[
x, x+1, x+2, \ldots, x+37
\]

The sum of these 38 numbers can be expressed using the formula for the sum of an arithmetic series:

\[
\text{Sum} = \text{Number of terms} \times \text{Average of the terms}
\]

In this case, the average of the 38 consecutive numbers is:

\[
\text{Average} = \frac{x + (x + 37)}{2} = \frac{2x + 37}{2}
\]

Thus, the sum of these 38 numbers is:

\[
\text{Sum} = 38 \times \frac{2x + 37}{2} = 19(2x + 37)
\]

We know that this sum equals 2071:

\[
19(2x + 37) = 2071
\]

To solve for \( x \), we start by dividing both sides by 19:

\[
2x + 37 = \frac{2071}{19}
\]

Calculating \( \frac{2071}{19} \):

\[
2071 \div 19 = 109
\]

So we can write:

\[
2x + 37 = 109
\]

Next, we solve for \( 2x \):

\[
2x = 109 - 37 = 72
\]

Now, divide by 2 to find \( x \):

\[
x = \frac{72}{2} = 36
\]

Thus, the first number is:

\[
\boxed{36}
\]

To verify, let’s calculate the sum of the 38 consecutive numbers starting from 36. The numbers are:

\[
36, 37, 38, \ldots, 73
\]

The number of terms is 38, and the sum is:

\[
\text{Sum} = \frac{38}{2} \times (36 + 73) = 19 \times 109 = 2071
\]

Since this matches the given sum of 2071, the calculation is confirmed. The first number is indeed \( \boxed{36} \).