The nth Term of an AP

The nth Term Formula

Writing out an AP term by term is fine for a few terms, but what if you need the 100th term? There's a direct formula.

The nth term (also called the general term) of an AP with first term $a$ and common difference $d$ is:

$$a_n = a + (n - 1)d$$

Here $n$ is the position of the term you want.

Why it works

The 1st term is $a$, the 2nd is $a + d$, the 3rd is $a + 2d$ — notice each term adds $d$ one fewer time than its position number. So the $n$th term adds $d$ exactly $(n-1)$ times.

Key Point: $a_n = a + (n-1)d$. Be careful — it's $(n-1)$, not $n$, multiplying $d$.

[Board Important] The single most common mistake is using $nd$ instead of $(n-1)d$. Always subtract 1 from the position first.

Finding a Particular Term

To find a specific term, just substitute the values of $a$, $d$, and $n$.

Worked outline: 10th term of $2, 7, 12, \dots$

• $a = 2$, $d = 5$, $n = 10$.
• $a_{10} = 2 + (10 - 1)(5) = 2 + 45 = 47$.

Finding $a$ or $d$ when a term is given

If you know one term and one of $a$ or $d$, the formula gives the other. For two given terms, you get two equations to solve for $a$ and $d$.

Key Point: The formula links four quantities $a, d, n, a_n$. Given any three, you can find the fourth.

[Board Important] When two terms are given (say $a_3 = 5$ and $a_7 = 17$), subtract to eliminate $a$: $a_7 - a_3 = 4d$, which gives $d$ immediately.

Which Term Equals a Given Value?

A common question: 'Which term of the AP is equal to (some number)?' Set $a_n$ equal to that number and solve for $n$.

Worked outline

Which term of $21, 18, 15, \dots$ is $-81$? Here $a = 21$, $d = -3$. $$a_n = 21 + (n-1)(-3) = -81 \Rightarrow 24 - 3n = -81 \Rightarrow n = 35.$$ So $-81$ is the 35th term.

Key Point: If solving gives a positive whole number $n$, the value is a term (the $n$th). If $n$ is not a whole number, the value is not a term of the AP.

[Board Important] 'Is 301 a term of $5, 11, 17, \dots$?' Solve $5 + (n-1)6 = 301$. If $n$ comes out fractional, the answer is 'no, not a term'.

The nth Term from the End

For a finite AP with last term $l$, the $n$th term counted from the end is found by treating the last term as the new first term and using $-d$ as the common difference:

$$\text{nth term from end} = l - (n - 1)d$$

Example

The AP $3, 8, 13, \dots, 253$ has $d = 5$. Its 3rd term from the end is $253 - (3-1)(5) = 253 - 10 = 243$.

Key Point: From the end, the common difference effectively becomes $-d$, so subtract $(n-1)d$ from the last term $l$.

[Board Important] Alternatively, you can find the total number of terms and convert to a position from the start — but the formula $l - (n-1)d$ is faster.

Solved Examples

Example 1: Find the 10th term

Find the 10th term of the AP $2, 7, 12, \dots$

Solution:

1. $a = 2$, $d = 5$, $n = 10$.
2. $a_{10} = 2 + 9(5) = 47$.

Final Answer: 47.

Takeaway: Substitute into $a_n = a + (n-1)d$.

Example 2: Which term is $-81$?

Which term of $21, 18, 15, \dots$ is $-81$?

Solution:

1. $a = 21$, $d = -3$. Set $a_n = -81$: $21 + (n-1)(-3) = -81$.
2. $24 - 3n = -81 \Rightarrow 3n = 105 \Rightarrow n = 35$.

Final Answer: The 35th term.

Takeaway: Set $a_n$ to the value and solve for $n$.

Example 3: Find the AP from two terms

The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.

Solution:

1. $a + 2d = 5$ and $a + 6d = 9$.
2. Subtract: $4d = 4 \Rightarrow d = 1$; then $a = 3$.
3. AP: $3, 4, 5, 6, \dots$

Final Answer: $3, 4, 5, 6, \dots$

Takeaway: Two terms give two equations; subtract to find $d$.

Example 4: Is 301 a term?

Is 301 a term of $5, 11, 17, 23, \dots$?

Solution:

1. $a = 5$, $d = 6$. Set $5 + (n-1)6 = 301$.
2. $6n - 1 = 301 \Rightarrow 6n = 302 \Rightarrow n = \dfrac{151}{3}$, not a whole number.

Final Answer: No, 301 is not a term.

Takeaway: A non-integer $n$ means the value is not in the AP.

Example 5: nth term from the end

Find the 11th term from the last term of the AP $10, 7, 4, \dots, -62$.

Solution:

1. $l = -62$, $d = -3$. nth term from end $= l - (n-1)d$.
2. $= -62 - (11 - 1)(-3) = -62 + 30 = -32$.

Final Answer: $-32$.

Takeaway: From the end, use $l - (n-1)d$ (here $d = -3$).

Example 6: Two-digit multiples of 3

How many two-digit numbers are divisible by 3?

Solution:

1. They form the AP $12, 15, 18, \dots, 99$ with $a = 12$, $d = 3$, $l = 99$.
2. $99 = 12 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow n - 1 = 29 \Rightarrow n = 30$.

Final Answer: 30 numbers.

Takeaway: Count terms by solving $l = a + (n-1)d$ for $n$.

Example 7: Find $d$ given two terms

If $a_5 = 19$ and $a_{12} = 47$ in an AP, find $a$ and $d$.

Solution:

1. $a + 4d = 19$ and $a + 11d = 47$.
2. Subtract: $7d = 28 \Rightarrow d = 4$; then $a = 19 - 16 = 3$.

Final Answer: $a = 3$, $d = 4$.

Takeaway: $a_{12} - a_5 = 7d$ gives $d$ at once.

Example 8: Find a later term using $d$

The 6th term of an AP is 12 and the common difference is 2. Find the 15th term.

Solution:

1. $a_6 = a + 5d = 12$ with $d = 2$: $a + 10 = 12 \Rightarrow a = 2$.
2. $a_{15} = 2 + 14(2) = 30$.

Final Answer: 30.

Takeaway: Find $a$ first, then substitute the desired $n$.

Example 9: Find $d$ from two given terms

In an AP, the 3rd term is 6 and the 7th term is 24. Find the common difference $d$.

Solution:

1. $a_3 = a + 2d = 6$; $a_7 = a + 6d = 24$.
2. Subtract: $4d = 18 \Rightarrow d = 4.5$.

Final Answer: $d = 4.5$.

Takeaway: Two term-equations always yield $d$ by subtraction.

Example 10: First negative term

Which is the first negative term of the AP $30, 27, 24, \dots$?

Solution:

1. $a = 30$, $d = -3$. We need $a_n < 0$: $30 + (n-1)(-3) < 0$.
2. $33 - 3n < 0 \Rightarrow n > 11$. So the first integer is $n = 12$.
3. $a_{12} = 30 + 11(-3) = -3$.

Final Answer: The 12th term, equal to $-3$.

Takeaway: Set up the inequality $a_n < 0$ and take the smallest integer $n$.