What Is The Nth Term Of An AP?
The nth term of an AP is the term sitting in position $n$ of an arithmetic progression, and it is given by $a_n = a + (n-1)d$. An arithmetic progression is a list of numbers where each term is a fixed amount more (or less) than the one before it. That fixed amount is the common difference, written $d$, and the starting value is the first term, written $a$.
Take the sequence $3, 7, 11, 15, 19, \ldots$ Here the first term is $a = 3$ and every term is $4$ more than the last, so $d = 4$. Instead of counting up term by term to reach, say, the 50th value, the nth-term formula jumps straight to it.
The value $a_n$ is also called the general term of the sequence, because a single expression describes every term at once. Feed it a position $n$ and it returns the number that lives there.
What Is The Formula For The Nth Term Of An AP?
The formula for the nth term of an AP is:
$$a_n = a + (n-1)d$$
Each symbol carries one job:
$a_n$ is the term you want, the one in position $n$.
$a$ is the first term, the value at position $1$.
$d$ is the common difference, found as $d = a_2 - a_1$.
$n$ is the position, a counting number ($1, 2, 3, \ldots$).
Before using the formula, pin down $a$ and $d$ from the sequence. The first term is simply the value you start with. The common difference is any term minus the term just before it, and for a proper AP that subtraction gives the same answer everywhere. A fuller treatment of that constant gap lives at common difference.
How Is The Nth Term Formula Derived?
The formula is not a rule to memorise blindly. It falls straight out of what an AP does: start at $a$, then add $d$ at every step.
Write the first few terms and watch the pattern:
$$a_1 = a$$ $$a_2 = a + d$$ $$a_3 = a + 2d$$ $$a_4 = a + 3d$$ $$a_5 = a + 4d$$
Table: How the number of added common differences tracks the position.
Position $n$ | Term written out | Copies of $d$ added |
|---|---|---|
1 | $a$ | 0 |
2 | $a + d$ | 1 |
3 | $a + 2d$ | 2 |
4 | $a + 3d$ | 3 |
$n$ | $a + (n-1)d$ | $n-1$ |
The number of $d$'s added is always one less than the position. At position $n$, the common difference has been added $(n-1)$ times, which gives:
$$a_n = a + (n-1)d$$
How Do You Find A Particular Term Of An AP?
To find a specific term, read off $a$ and $d$, then substitute the position into the formula.
Example 1: Find the 20th term of the AP $3, 7, 11, 15, \ldots$
Here $a = 3$, $d = 4$, and $n = 20$. Substitute:
$$a_{20} = 3 + (20-1)\cdot 4$$ $$a_{20} = 3 + 19 \cdot 4$$ $$a_{20} = 3 + 76 = 79$$
Final answer: the 20th term is $79$.
Notice the arithmetic on the bracket happened first: $20 - 1 = 19$, then $19 \times 4$. Doing $20 \times 4$ instead is the classic slip, and the common-mistakes section below returns to it.
How Do You Find N When You Know The Term?
Set the nth-term formula equal to the known value and solve for $n$. If $n$ comes out as a whole number, the value belongs to the sequence; if not, it does not.
Example 2: Which term of the AP $5, 8, 11, \ldots$ is $302$?
Here $a = 5$ and $d = 3$. Set $a_n = 302$:
$$5 + (n-1)\cdot 3 = 302$$ $$(n-1)\cdot 3 = 297$$ $$n - 1 = 99$$ $$n = 100$$
Final answer: $302$ is the 100th term.
Because $n = 100$ is a whole number, $302$ genuinely sits in the sequence. Had the division left a fraction, the honest answer would be that $302$ is not a term at all.
How Do You Find The Common Difference Or First Term?
When two terms are known, write the formula for each, then subtract one equation from the other. The subtraction cancels $a$ and leaves an equation in $d$ alone.
Example 3: In an AP the 3rd term is $12$ and the 7th term is $24$. Find $d$ and $a$.
Write both terms with the formula:
$$a_3 = a + 2d = 12$$ $$a_7 = a + 6d = 24$$
Subtract the first from the second:
$$4d = 12 \quad\Rightarrow\quad d = 3$$
Put $d = 3$ back into $a + 2d = 12$:
$$a + 6 = 12 \quad\Rightarrow\quad a = 6$$
Final answer: the common difference is $3$ and the first term is $6$.
This subtract-to-cancel move is worth keeping. Any two terms of an AP give you $d$ the moment you subtract their equations, because the gap between positions is all that survives.
What Is The Nth Term From The End Of An AP?
The nth term counted from the end of a finite AP uses the last term $l$ and steps backward, so the common difference is subtracted:
$$(\text{nth term from the end}) = l - (n-1)d$$
Counting from the end simply flips the direction of travel. Each step back lowers the value by $d$ instead of raising it, so the sign in front of the bracket flips from plus to minus.
Example 4: Find the 4th term from the end of $3, 7, 11, \ldots, 99$.
Here the last term is $l = 99$ and $d = 4$, with $n = 4$:
$$99 - (4-1)\cdot 4 = 99 - 12 = 87$$
Final answer: the 4th term from the end is $87$.
Check it by hand: from the end the terms run $99, 95, 91, 87$, and the fourth is indeed $87$. There is also an equivalent way that uses the total number of terms $m$: the nth term from the end equals the $(m-n+1)$th term from the start.
How Is The Nth Term Of An AP Different From A GP?
An AP and a GP both build a sequence from a first term, but an AP adds a fixed amount at each step while a GP multiplies by a fixed ratio. That single difference changes the whole formula.
Table: The nth term in an arithmetic progression versus a geometric progression.
Arithmetic progression (AP) | Geometric progression (GP) | |
|---|---|---|
Step rule | Add the common difference $d$ | Multiply by the common ratio $r$ |
nth term | $a_n = a + (n-1)d$ | $a_n = a,r^{,n-1}$ |
Example | $2, 5, 8, 11, \ldots$ ($d = 3$) | $2, 6, 18, 54, \ldots$ ($r = 3$) |
Growth shape | Straight-line, steady | Curved, speeds up (or shrinks) fast |
The shared piece is the exponent and the bracket: both formulas use $(n-1)$, because both count steps between terms rather than the terms themselves. Reaching for $r$ when the sequence actually adds, or for $d$ when it multiplies, is a frequent mix-up. The parallel case for a multiplying sequence is spelled out at nth term of a GP.
Why Does The Nth Term Formula Use (n − 1)?
The $(n-1)$ is where most marks are won or lost, so it deserves a plain answer rather than a shrug at the formula sheet. The reason comes down to counting jumps instead of terms.
The first term needs no jump. It is already there at position $1$, so zero copies of $d$ have been added yet.
Each move adds exactly one $d$. Going from position $1$ to position $n$ takes $n-1$ moves, so $d$ is added $n-1$ times.
Terms and gaps are different counts. A row of $n$ terms has only $n-1$ gaps between them, the same way a fence with $n$ posts has $n-1$ panels.
So the formula reads position $n$ as the first term plus $(n-1)$ jumps of size $d$. Miss the subtraction and you add one jump too many, which lands you on the $(n+1)$th term every single time.
Who Discovered Arithmetic Progressions?
Arithmetic progressions are old, far older than the modern notation. Rules for adding evenly spaced numbers appear on Babylonian clay tablets, and Indian mathematicians turned those patterns into general methods more than 1,400 years ago.
Two other names shaped the same ground:
Brahmagupta (598–668 CE, India) gave general rules for summing arithmetic series in the Brahmasphutasiddhanta, extending the work of the generation before him.
Carl Friedrich Gauss (1777–1855, Germany) is said to have added $1$ to $100$ in seconds as a schoolboy by pairing the first and last terms, the same first-plus-last insight that powers arithmetic-series formulas today.
Where Is The Nth Term Of An AP Used In The Real World?
Any quantity that changes by a fixed amount at each step is an AP, and its nth term answers "what is the value at step $n$" without walking through every step.
Personal finance: money set aside at a fixed amount each month grows as an AP, and the nth term gives the total contributed by any month. Simple interest builds the same way.
Architecture and seating: stadium and theatre rows that grow by a set number of seats form an AP, so the nth term gives the seats in any row.
Construction and stacking: logs, pipes, or cans stacked with one fewer per rising row are an AP counted from the top down.
Salaries and planning: a wage with a fixed yearly raise is an AP, and the nth term gives the pay in any future year.
Depreciation: an asset losing a fixed amount of value each year (straight-line depreciation) is a decreasing AP, and the nth term gives its worth in year $n$.
One short formula reaches across finance, design, engineering, and accounting. The pattern of steady, equal steps shows up wherever change is predictable.
What Are The Most Common Nth Term Of An AP Mistakes?
These four errors cause most lost marks on AP questions, matching the student errors documented across published error studies and revision guides.
Using $n$ instead of $(n-1)$.
Where it slips in:
A student finds the 20th term by adding $20$ copies of $d$ to the first term, when only $19$ jumps separate term $1$ from term $20$.
Don't do this:
Do not multiply $d$ by the position number itself. That adds one jump too many.
The correct way:
Always subtract one first. Compute $a_n = a + (n-1)d$, so for the 20th term of $3, 7, 11, \ldots$ use $3 + 19\cdot 4 = 79$, not $3 + 20\cdot 4$.
Getting the sign of the common difference wrong.
Where it slips in:
In a falling sequence like $50, 44, 38, \ldots$, a student writes $d = 6$ and the terms start climbing instead of dropping.
Don't do this:
Do not assume $d$ is positive. A decreasing AP has a negative common difference.
The correct way:
Compute $d = a_2 - a_1$ with the sign. Here $d = 44 - 50 = -6$, so $a_n = 50 + (n-1)(-6)$.
Confusing the first term with the common difference.
Where it slips in:
A student swaps $a$ and $d$, plugging the gap into the starting slot or the starting value into the jump.
Don't do this:
Do not treat $a$ and $d$ as interchangeable. The first term is where the sequence begins; the common difference is how far each step moves.
The correct way:
Label them before substituting. Write $a = \text{(first value)}$ and $d = \text{(one term minus the previous)}$ off to the side, then fill the formula.
Mixing up the AP formula with the GP formula.
Where it slips in:
A student reaches for $a,r^{n-1}$ on a sequence that adds a constant, or uses $a + (n-1)d$ on one that multiplies.
Don't do this:
Do not choose the formula by habit. Check whether consecutive terms differ by a fixed amount or by a fixed multiple first.
The correct way:
Subtract two neighbouring terms and divide two neighbouring terms. If the differences match, it is an AP and you use $a + (n-1)d$; if the ratios match, it is a GP.
Practice Problems On The Nth Term Of An AP
Work each one, then check against the answer beside it.
Find the 15th term of $2, 6, 10, 14, \ldots$
(Answer: $a = 2$, $d = 4$, so $a_{15} = 2 + 14\cdot 4 = 58$.)Which term of $7, 13, 19, \ldots$ is $205$?
(Answer: $7 + (n-1)6 = 205 \Rightarrow n = 34$, the 34th term.)The 5th term of an AP is $19$ and the 9th term is $35$. Find $a$ and $d$.
(Answer: subtracting gives $4d = 16$, so $d = 4$ and $a = 3$.)An AP has first term $100$ and common difference $-7$. Find the 12th term.
(Answer: $100 + 11(-7) = 23$.)Find the 6th term from the end of $5, 8, 11, \ldots, 95$.
(Answer: $95 - (6-1)\cdot 3 = 80$.)Is $100$ a term of the AP $3, 7, 11, 15, \ldots$?
(Answer: $3 + (n-1)4 = 100 \Rightarrow n - 1 = 24.25$, not a whole number, so no.)
Where Should You Go Next After The Nth Term Of An AP?
The nth term is one door into a whole family of sequence tools, and a few natural next steps open from here.
Sum of n terms of an AP. Once you can find any single term, the next question is adding a stretch of them at once.
Arithmetic progressions. Step back to the full picture of first term, common difference, terms, and sums together.
Nth term of a GP. See how the same position logic behaves when a sequence multiplies instead of adds.
Sequences. Zoom out to how mathematicians describe ordered lists of numbers in general.
If your child is building these foundations, a live Bhanzu trainer teaches the nth term starting from the "why" behind the $(n-1)$ in the Bhanzu algebra program.
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