General ($n$th) term of an AP

If you know the first term and the common difference, you should be able to write down the $20$th term without listing all the others. The closed-form formula for the $n$th term makes this immediate.

The formula

Let an AP have first term $a$ and common difference $d$. Then the $n$th term is $\boxed{a_n = a + (n - 1) d.}$

Derivation

Start with $a_1 = a$. Each step adds $d$: $a_2 = a + d,$ $a_3 = a_2 + d = a + 2 d,$ $a_4 = a_3 + d = a + 3 d,$ $\ldots$ $a_n = a + (n - 1) d.$

The exponent on $d$ is $(n - 1)$ because the first term itself doesn't "step". After the $1$st term, we add $d$ exactly $(n - 1)$ times to reach the $n$th.

Using the formula

Three classical patterns appear repeatedly.

Pattern 1: Find a specific term. Given $a, d, n$, just plug in.

Pattern 2: Find which term is a given value. Given $a, d$, and a target value $T$, set $a + (n - 1) d = T$ and solve for $n$. The answer should be a positive integer; otherwise $T$ is not actually a term of the AP.

Pattern 3: Find $a$ and $d$ from two terms. If the $m$th term is $p$ and the $n$th term is $q$, you have $a + (m - 1) d = p, \quad a + (n - 1) d = q.$ Subtract to get $(m - n) d = p - q$, hence $d = (p - q)/(m - n)$. Then back-substitute for $a$.

Worked examples

Example 1. Find the $10$th term of $2, 7, 12, 17, \ldots$.

$a = 2, d = 5$. $a_{10} = 2 + 9 \cdot 5 = 47$.

Example 2. Which term of the AP $3, 8, 13, \ldots$ is $78$?

$a = 3, d = 5$. Set $3 + (n - 1) \cdot 5 = 78 \Rightarrow 5(n - 1) = 75 \Rightarrow n = 16$. The $16$th term.

Example 3. The $4$th term of an AP is $11$ and the $8$th is $23$. Find $a$ and $d$.

$a + 3 d = 11$, $a + 7 d = 23$. Subtract: $4 d = 12 \Rightarrow d = 3$. Then $a = 11 - 9 = 2$.

Example 4. Find the number of terms of the AP $7, 13, 19, \ldots, 205$.

$a = 7, d = 6, \ell = 205$. Set $a + (n - 1) d = 205 \Rightarrow 7 + 6(n - 1) = 205 \Rightarrow n - 1 = 33 \Rightarrow n = 34$.

Example 5. Find the $20$th term from the end of the AP $3, 8, 13, \ldots, 253$.

Method: read the AP in reverse. Reversed AP starts at $253$ with common difference $-5$. Its $20$th term is $253 + 19 \cdot (-5) = 253 - 95 = 158$.

Alternative: total terms $n$ satisfies $3 + (n - 1) \cdot 5 = 253 \Rightarrow n = 51$. The $20$th from the end is the $51 - 20 + 1 = 32$nd from the start: $3 + 31 \cdot 5 = 158$. ✓

Try it yourself

1. Find the $25$th term of $5, 10, 15, 20, \ldots$.
2. Find $a_{15}$ if $a_3 = 7$ and $a_{10} = 28$.
3. Which term of the AP $21, 18, 15, \ldots$ is $-81$? Is any term zero?
4. The $17$th term of an AP exceeds its $10$th term by $7$. Find the common difference.
5. How many three-digit numbers are divisible by $7$? (AP $105, 112, \ldots, 994$.)
6. Find the $11$th term from the end of the AP $10, 7, 4, \ldots, -62$.
7. If the $9$th term of an AP is zero, prove that its $29$th term is twice its $19$th term.
8. The first term of an AP is $5$, the last term is $45$, and the sum is $400$. Find the number of terms.
9. Find the $20$th term of an AP whose $5$th term is $13$ and $11$th term is $25$.
10. If $a_n = 2 n + 3$, find $a, d$, and the $30$th term.

Pitfalls / Insight

• The $(n-1)$ in the formula is easy to misremember. Test on $n = 1$ to anchor: $a_1$ should give $a$.
• $n$ must be a positive integer. If solving for $n$ gives a fraction or a negative, $T$ is not a term of the AP.
• For "$k$th from the end", use the reverse-AP trick or compute $n - k + 1$.

Insight. Two pieces of information about an AP (e.g., two specific terms) always pin down both $a$ and $d$. After that, everything else is plug-and-chug.