TMUA practice: Arithmetic Series, problem 3

The sum of the first $n$ positive integers is $\;S_{n} = \dfrac{n(n + 1)}{2}$.

Set this equal to $210$:

Apply the quadratic formula or factorise. Look for two integers multiplying to $-420$ and adding to $1$: $\;21 \cdot 20 = 420$, $\;21 - 20 = 1$ ✓. So

$n^{2} + n - 420 \;=\; (n - 20)(n + 21) \;=\; 0.$

The positive root is $n = 20$. (The other root $n = -21$ is rejected as $n$ must be a positive integer.)

Verify: $S_{20} = 20 \cdot 21 / 2 = 210$ ✓.

This is option C.

Why the other options fail.

• A, $10$: $S_{10} = 55$, not $210$.
• B, $15$: $S_{15} = 120$, not $210$.
• D, $21$: $S_{21} = 231$, not $210$. (This is the magnitude of the discarded negative root.)
• E, $30$: $S_{30} = 465$, much too large.

The lesson: $S_{n}$ for $1 + 2 + \dots + n$ grows roughly like $n^{2}/2$, so to reach $210$ we expect $n \approx \sqrt{420} \approx 20.5$ — consistent with the exact answer $n = 20$.