4. The sum of the first $n$ natural numbers is given by the formula $\frac{n(n+1)}{2}$. Find the sum of the natural numbers between $60$ and $400$ inclusive in multiples of $5$.

Key concept: This is an arithmetic series: the numbers increase by 5, so the sum can be found with the formula for the sum of an arithmetic sequence.

Step by step

Step 1: List the first and last terms

The multiples of 5 between 60 and 400 inclusive are:

$60, 65, 70, \dots, 400$

So:

• first term $a_1 = 60$
• last term $a_n = 400$
• common difference $d = 5$

Step 2: Find the number of terms

Use

$n = \frac{400 - 60}{5} + 1 = 68 + 1 = 69$

Step 3: Use the arithmetic series sum formula

$S_n = \frac{n(a_1 + a_n)}{2}$

Substitute the values:

$S_{69} = \frac{69(60 + 400)}{2} = \frac{69 \cdot 460}{2} = 69 \cdot 230 = 15870$

Answer

$\boxed{15870}$

Pitfall alert

A frequent error is forgetting to include both endpoints. Since 60 and 400 are multiples of 5, both must be counted. Another mistake is using the formula for the sum of the first $n$ natural numbers instead of the arithmetic-series sum formula.

Try different conditions

If the interval were open instead of inclusive, you would exclude 60 and 400 and start from 65 and end at 395. The method would be the same, but the first term, last term, and number of terms would change.

Further reading

arithmetic series, inclusive endpoints, common difference