nth term of an AP

In an AP with first term $a$ and common difference $d$, the terms are $a,\ a+d,\ a+2d,\ a+3d,\ \ldots$ Look at how many copies of $d$ each term carries: the 1st term has $0$ copies, the 2nd has $1$, the 3rd has $2$ — the pattern is always one fewer copy of $d$ than the term's position. This gives the central formula of the chapter:

where: $a_n$ is the value of the $n$th term, $a$ is the first term of the AP, $d$ is the common difference (found as $d = a_2 - a_1$, or any later term minus the one before it), and $n$ is a positive integer giving the position of the term counted from the start. The same formula runs in more than one direction. Given $a$, $d$ and $n$, you compute $a_n$ directly. Given $a_n$, $a$ and $d$, you can instead solve for $n$ — this tells you which position a stated value occupies, or reveals that the value never appears in the AP at all (if the resulting $n$ is not a positive whole number). And given any two terms of an AP, you can write the formula once for each term to get two equations, then solve them together for $a$ and $d$. A closely related idea is the last term of a finite AP with $n$ terms: if the AP is $a, a+d, \ldots, l$, the last term is simply $l = a + (n-1)d$ — the same formula, with the final position substituted for $n$.

Step 1. Find the 15th term of the AP $7, 11, 15, 19, \ldots$ Identify $a$ and $d$: here $a = 7$, and $d = 11 - 7 = 4$ (check: $15 - 11 = 4$, $19 - 15 = 4$, consistent throughout).

Step 2. We want $a_{15}$, so $n = 15$. Substitute into $a_n = a + (n-1)d$: $a_{15} = 7 + (15-1)\times 4 = 7 + 14\times 4 = 7 + 56 = \boxed{63}$.

Step 3. Now run the formula the other way. Which term of the AP $3, 8, 13, 18, \ldots$ is $78$? Here $a = 3$, $d = 8 - 3 = 5$, and $a_n = 78$; the unknown this time is $n$. $78 = 3 + (n-1)\times 5 \Rightarrow 75 = (n-1)\times 5 \Rightarrow n - 1 = 15 \Rightarrow n = 16$. Check: $a_{16} = 3 + 15\times 5 = 3 + 75 = 78$. ✓ So $78$ is the 16th term.

Competitive stretch

Step 1. Suppose only two terms are given: $a_5 = 18$ and $a_9 = 30$, and $a$, $d$ are both unknown. Write the $n$th-term formula for each: $a_5 = a + 4d = 18$ and $a_9 = a + 8d = 30$.

Step 2. Subtract the first equation from the second to eliminate $a$: $(a+8d) - (a+4d) = 30 - 18 \Rightarrow 4d = 12 \Rightarrow d = 3$.

Step 3. Substitute back: $a + 4(3) = 18 \Rightarrow a = 6$. So $a = 6$ and $d = 3$ — the same technique from Pair of Linear Equations, applied to two AP terms.

• Using $n$ instead of $(n-1)$. The formula needs $(n-1)$ copies of $d$, not $n$ copies; this off-by-one slip shifts every answer by one common difference.
• Substituting into the wrong slot. Before computing, write down clearly which of $a, d, n, a_n$ you already know and which one you are solving for.
• Losing the sign of a negative $d$. For a decreasing AP, carry the negative sign through the whole calculation of $(n-1)d$ rather than dropping it midway.
• Assuming every number is a term. When checking whether a number is a term of the AP, solving for $n$ must give a positive integer; a fractional or negative $n$ means the number is not actually in that AP.

Ask the tutor: If rearranging $a_n = a + (n-1)d$ to solve for $n$, $a$, or $d$ feels shaky, ask the AI tutor to walk through the algebra one line at a time with a fresh AP.