The First Negative Term of an AP

For a decreasing AP, the terms eventually go negative. Find where by writing the nth term, setting it $< 0$, and solving for $n$, then round to a whole number and check, because $n$ counts terms.

The arithmetic progression $20, 17, 14, \dots$ has first term $20$ and common difference $-3$. Find the first term of the progression that is negative. [4]

The working

Step 1, write the nth term with $u_n = a + (n - 1)d$, $a = 20$, $d = -3$: $u_n = 20 + (n - 1)(-3) = 20 - 3n + 3 = 23 - 3n \quad \text{(M1)}$

Step 2, set the term $< 0$: $23 - 3n < 0 \;\Rightarrow\; 23 < 3n \;\Rightarrow\; n > 7.67 \quad \text{(M1, A1)}$

Step 3, round up and identify the term. Since $n$ is a whole number and $n > 7.67$, the first negative term is the 8th ($n = 8$): $u_8 = 23 - 3(8) = 23 - 24 = -1 \quad \text{(A1)}$

Check: $u_7 = 23 - 21 = 2 > 0$ (still positive), confirming the 8th term is the first negative one.

Where the marks are won and lost

• Simplify the nth term to $23 - 3n$ carefully: $20 + (n-1)(-3) = 20 - 3n + 3$. The $+3$ from $-(n-1)(3)$ expansion is easy to miss.
• Round up ($n > 7.67 \Rightarrow n = 8$): the term must be past the boundary. Rounding down gives $n = 7$, which is still positive.
• The confirming check ($u_7 > 0$, $u_8 < 0$) secures the answer.

Common mistakes

• Sign or expansion errors giving the wrong nth-term formula.
• Rounding $7.67$ down to $7$.
• Finding the value where the term equals zero and stopping, without identifying the first negative term.

Full method: Arithmetic Progressions notes. Topic home: Series pillar.