How Many AP Terms Are Needed to Exceed a Value

This question ends where a quadratic-inequality question begins: the sum formula produces a quadratic in $n$, and “how many terms” forces $n$ to be a whole number. The finish is a decision about rounding, plus a check.

The arithmetic progression $3, 7, 11, 15, \dots$ has first term $3$ and common difference $4$. Find the least number of terms for which the sum first exceeds $200$. [5]

The working

Step 1, write the sum $S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]$ with $a = 3$, $d = 4$: $S_n = \frac{n}{2}\big[6 + 4(n - 1)\big] = \frac{n}{2}\big[4n + 2\big] = n(2n + 1) = 2n^2 + n \quad \text{(M1, A1)}$

Step 2, form the inequality $S_n > 200$: $2n^2 + n > 200 \;\Rightarrow\; 2n^2 + n - 200 > 0 \quad \text{(M1)}$

Step 3, find the boundary by solving $2n^2 + n - 200 = 0$ with the formula: $n = \frac{-1 + \sqrt{1 + 1600}}{4} = \frac{-1 + \sqrt{1601}}{4} = \frac{-1 + 40.01}{4} \approx 9.75 \quad \text{(A1)}$

(The negative root is discarded, $n$ must be positive.)

Step 4, round up and check. Since $n > 9.75$ and $n$ is a whole number, the least value is $n = 10$. Verify: $S_{9} = 2(81) + 9 = 171 \ (\le 200), \qquad S_{10} = 2(100) + 10 = 210 \ (> 200)$ $\Rightarrow \textbf{10 terms} \quad \text{(A1)}$

Where the marks are won and lost

• Simplifying $S_n$ to $2n^2 + n$ cleanly makes the inequality manageable. Leaving it as $\frac{n}{2}[4n+2]$ and solving is possible but messier.
• Round up, not down. The sum must exceed $200$; $n = 9.75$ rounds to $10$ because $9$ is not yet enough.
• The check ($S_9$ vs $S_{10}$) confirms the boundary and catches rounding errors, examiners reward it and it protects the final mark.

Common mistakes

• Rounding $9.75$ down to $9$.
• Treating $n$ as continuous and giving $9.75$ as the answer.
• Slips in the quadratic formula, especially $\sqrt{1601}$ and the division by $2a = 4$.

Full method: Arithmetic Progressions notes. See also Quadratic Inequalities. Topic home: Series pillar.