Why write out thirty terms when one formula jumps straight to the one you want? The nth term $a_n=a+(n-1)d$ finds any term, tells you whether a number belongs to an AP, and works just as well counting back from the end.
Meera's savings jar from Section 5.1 gets ₹ $200$ in January, ₹ $250$ in February, ₹ $300$ in March, and so on. How much does she put in in December? You could keep adding $50$ eleven more times — or you could notice that December is the $12$th month, so it is $11$ steps of ₹ $50$ after January: ₹ $200+11\times50=$ ₹ $750$. The same idea answers questions that would take far too long by listing: the number of seats in the $25$th row of a stand, the fare for a $38$ km taxi ride, the first month in which a saving goes above ₹ $500$.
In this section we turn "count the steps" into a formula for the $n$th term of an AP. We use it to find any term, to decide whether a given number is a term (and which one), to rebuild a whole AP from just two of its terms, and to count terms backwards from the end of a finite AP. The formula is short, but it is used in almost every AP question you will meet.
A formula for the nth term
Write the terms of an AP with first term $a$ and common difference $d$ next to their positions, and count how many $d$'s each one contains:
Position $n$
$1$
$2$
$3$
$4$
$\dots$
$n$
Term $a_n$
$a$
$a+d$
$a+2d$
$a+3d$
$\dots$
$a+(n-1)d$
Steps of $d$ from $a$
$0$
$1$
$2$
$3$
$\dots$
$n-1$
To reach the $n$th term you start at $a$ and take one step fewer than the position number — just as there are $3$ gaps between $4$ fence posts.
📐 Theorem 5.2 — The nth term of an AP
In an AP with first term $a$ and common difference $d$, the $n$th term is
$$a_n=a+(n-1)d,\qquad n=1,2,3,\dots$$
$a_n$ is also called the general term of the AP. In particular, a finite AP with $m$ terms has last term $l=a_m=a+(m-1)d$.
⚠️ Common mistakes
1. Using $n$ instead of $n-1$. The $20$th term of $5, 9, 13, \dots$ is $5+19\times4=81$, not $5+20\times4=85$ (that is the $21$st term).
2. Losing the sign of $d$. For $10, 7, 4, \dots$, $d=-3$, so $a_{15}=10+14\times(-3)=-32$. Put $d$ in brackets when it is negative.
3. Accepting a non-whole $n$. Positions are $1, 2, 3, \dots$. If solving $a+(n-1)d=k$ gives $n=50.3$ or $n=-2$, then $k$ is not a term of the AP.
✏️ Example 1 — Finding a term
(a) Find the $20$th term of $5, 9, 13, \dots$ (b) Find the $15$th term of $10, 7, 4, \dots$ (c) Find the $n$th term of $-4, -1, 2, \dots$
Check In (c), $n=1, 2, 3$ give $3-7=-4$, $6-7=-1$, $9-7=2$ ✓ — the formula reproduces the given terms.
✏️ Example 2 — Is it a term? Which term?
(a) Which term of $3, 8, 13, \dots$ is $78$? (b) Is $301$ a term of $5, 11, 17, \dots$? (c) Which term of $21, 18, 15, \dots$ is $-81$?
💡 Show solution▶
(a) $a=3$, $d=5$. Put $a_n=78$: $\;3+(n-1)\times5=78$, so $(n-1)\times5=75$, $n-1=15$, $n=16$. $78$ is the $16$th term.
(b) $a=5$, $d=6$. $\;5+(n-1)\times6=301$ gives $n-1=\tfrac{296}{6}=49\tfrac13$. This is not a whole number, so $301$ is not a term. (Indeed $a_{50}=299$ and $a_{51}=305$; $301$ falls between them.)
Each bar is a term of the AP. The highlighted bar $a_n$ is built from the first term $a$ (blue) plus $n-1$ red blocks, each of size $d$ — count them! Press Play to let $n$ grow, or set $a$, $d$ and $n$ yourself. With a negative $d$ the blocks are stacked downwards.
347
the AP—
blocks of $d$—
$a_n=a+(n-1)d$—
✏️ Example 3 — An AP from two of its terms
The $3$rd term of an AP is $12$ and the $10$th term is $40$. Find the AP, its $20$th term, and the term that equals $100$.
💡 Show solution▶
Two equations $a_3=a+2d=12$ and $a_{10}=a+9d=40$.
Eliminate $a$ Subtract the first from the second: $7d=28$, so $d=4$. Then $a=12-2\times4=4$. The AP is $4, 8, 12, 16, \dots$
20th term $a_{20}=4+19\times4=80$.
Which term is 100? $4+(n-1)\times4=100$ gives $n-1=24$, $n=25$.
Check $a_{10}=4+9\times4=40$ ✓. Notice the shortcut: from the $3$rd to the $10$th term is $7$ steps, and the terms rose by $28$, so each step is $4$.
Using the nth term in problems
The formula $a_n=a+(n-1)d$ links four quantities: $a$, $d$, $n$ and $a_n$. Every question gives you some of them and asks for another. Word problems simply hide the AP inside a story.
🧭 Problem-solving strategy — nth term problems
Write the AP (or its first few terms) and read off $a$ and $d$. In a story, decide carefully what the first term stands for.
Decide which of $a$, $d$, $n$, $a_n$ is unknown, and write $a_n=a+(n-1)d$ with the known values substituted.
Solve. If you are finding $n$, it must be a positive whole number. For "the first term greater than (or less than) $k$", solve the inequality and take the smallest whole number that works.
Check by substituting back, and answer in the words (and units) of the question.
✏️ Example 4 — Counting multiples
How many three-digit numbers are divisible by $7$?
💡 Show solution▶
The AP The smallest three-digit multiple of $7$ is $105$ ($=7\times15$), and the largest is $994$ ($=7\times142$, since $7\times143=1001$ has four digits). The multiples form the AP $105, 112, 119, \dots, 994$ with $a=105$, $d=7$, $l=994$.
Count $994=105+(n-1)\times7$, so $(n-1)\times7=889$, $n-1=127$, $n=128$.
Check The multiples are $7\times15$ up to $7\times142$, and there are $142-15+1=128$ of them ✓.
✏️ Example 5 — A savings plan
Arjun saves ₹ $120$ in the first month, and every month he saves ₹ $15$ more than in the month before. (a) How much does he save in the $12$th month? (b) In which month does he save ₹ $405$? (c) In which month does his saving first go above ₹ $500$?
💡 Show solution▶
The AP Monthly savings ₹ $120, 135, 150, \dots$: $a=120$, $d=15$.
(a) $a_{12}=120+11\times15=120+165=285$. He saves ₹ $285$.
(b) $120+(n-1)\times15=405$ gives $(n-1)\times15=285$, $n-1=19$, $n=20$: in the $20$th month.
(c) Inequality We need $120+(n-1)\times15\gt500$, i.e. $(n-1)\times15\gt380$, $n-1\gt25\tfrac13$. The smallest whole number $n-1$ that works is $26$, so $n=27$.
(c) Check $a_{26}=120+25\times15=495$ (not above $500$) and $a_{27}=120+26\times15=510$ (above $500$) ✓. Answer: the $27$th month, when he saves ₹ $510$.
Counting from the end of a finite AP
Sometimes a question counts backwards: "the $10$th term from the end". Read a finite AP backwards and you get another AP. Its first term is the last term $l$, and because you are walking the other way, its common difference is $-d$. For example, $3, 8, 13, \dots, 253$ read backwards is $253, 248, 243, \dots, 3$.
📐 Theorem 5.3 — The nth term from the end
In a finite AP with last term $l$ and common difference $d$, the $n$th term from the end is
$$l-(n-1)d.$$
If the AP has $m$ terms, the $n$th term from the end is the same as the $(m-n+1)$th term from the beginning.
Why it is true. The reversed AP has first term $l$ and common difference $-d$, so by Theorem 5.2 its $n$th term is $l+(n-1)(-d)=l-(n-1)d$. Counting positions: the last term is the $m$th from the start, the second-last is the $(m-1)$th, and the $n$th from the end is the $(m-n+1)$th.
Interactive 2 — Counting from Both Ends
A finite AP with $m$ terms is laid out as tiles. The blue arrow counts $k$ terms from the start, the red arrow counts $k$ terms from the end. Press Play to move $k$ along. Watch two things: the $k$th term from the end is always the $(m-k+1)$th from the start, and the two highlighted terms always add up to the same number, $a+l$.
35103
$k$th from the start—
$k$th from the end—
their sum—
✏️ Example 6 — The 10th term from the end
Find the $10$th term from the end of the AP $3, 8, 13, \dots, 253$.
Method 2 — count the terms $253=3+(m-1)\times5$ gives $m-1=50$, $m=51$ terms. The $10$th from the end is the $(51-10+1)$th $=42$nd term from the start.
Method 2 — the term $a_{42}=3+41\times5=3+205=208$ ✓ — both methods agree.
Look ahead The $10$th term from the start is $a_{10}=48$, and $48+208=256=3+253=a+l$. Pairing terms from the two ends like this is the key to adding up an AP in Section 5.3.
Exercises
I
Find the indicated term.
1 the $10$th term of $2, 7, 12, \dots$
2 the $18$th term of $30, 26, 22, \dots$
3 the $25$th term of $-3, -\tfrac12, 2, \dots$
4 the $12$th term of $\sqrt2, 3\sqrt2, 5\sqrt2, \dots$
5 the $n$th term of $7, 11, 15, \dots$
6 the $n$th term of $5, 2, -1, \dots$
7 the $11$th term of $1.6, 1.9, 2.2, \dots$
8 the $31$st term of an AP whose $11$th term is $38$ and $16$th term is $73$
5 Which term of $\tfrac56, 1, \tfrac76, \dots$ is $3$?
6 How many terms does the AP $7, 13, 19, \dots, 205$ have?
7 Which is the first negative term of $121, 117, 113, \dots$?
8 Which term of $3, 10, 17, \dots$ is $84$ more than its $13$th term?
Check: type $a$, $d$ and a number $k$ — is $k$ a term, and which one?
solve a + (n − 1)d = k—
nearest terms—
💡 Show solutions▶
1 $3+(n-1)\times12=135$: $\;n-1=11$, $n=12$. The $12$th term.
2 $84+(n-1)(-4)=0$: $\;n-1=21$, $n=22$. The $22$nd term.
3 $3+(n-1)\times4=184$: $\;n-1=\tfrac{181}4=45.25$, not whole. No ($a_{46}=183$, $a_{47}=187$).
4 $11+(n-1)(-3)=-150$: $\;n-1=\tfrac{161}3=53\tfrac23$, not whole. No.
5 $d=\tfrac16$: $\;\tfrac56+(n-1)\times\tfrac16=3$ gives $(n-1)\times\tfrac16=\tfrac{13}6$, $n-1=13$, $n=14$. The $14$th term.
6 $7+(n-1)\times6=205$: $\;n-1=33$, so there are $34$ terms.
7 $121+(n-1)(-4)\lt0$ means $n-1\gt30.25$, so $n=32$: $\;a_{32}=121-124=-3$. (Check $a_{31}=1\gt0$.) The $32$nd term, $-3$.
8 $a_{13}=3+12\times7=87$, so we want $87+84=171$: $\;3+(n-1)\times7=171$, $n-1=24$, $n=25$. (Shortcut: $84\div7=12$ more steps after the $13$th term.)
III
Finding the AP from given information.
1 The $4$th term of an AP is $11$ and the $9$th term is $31$. Find the AP and its $20$th term.
2 The $3$rd term of an AP is $16$, and the $7$th term exceeds the $5$th term by $12$. Find the first four terms.
3 The $17$th term of an AP exceeds its $10$th term by $7$, and its first term is $2$. Find the $30$th term.
4 In an AP, $a_4+a_8=24$ and $a_6+a_{10}=44$. Find the first three terms.
5 Two APs have the same common difference. Their first terms are $-1$ and $-8$. Find the difference between their $4$th terms, and between their $100$th terms. What do you notice?
6 In an AP, $7$ times the $7$th term equals $11$ times the $11$th term. Show that the $18$th term is $0$.
7 For what value of $n$ are the $n$th terms of the APs $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?
2 $a_7-a_5=2d=12$, so $d=6$; $\;a+2d=16$ gives $a=4$. The terms are $4, 10, 16, 22$.
3 $a_{17}-a_{10}=7d=7$, so $d=1$. $\;a_{30}=2+29\times1=31$.
4 $(a+3d)+(a+7d)=2a+10d=24$ and $(a+5d)+(a+9d)=2a+14d=44$. Subtract: $4d=20$, $d=5$; then $2a=24-50=-26$, $a=-13$. Terms $-13, -8, -3$. (Check: $a_4=2$, $a_8=22$, sum $24$ ✓.)
5 With common difference $d$: $(-1+3d)-(-8+3d)=7$ and $(-1+99d)-(-8+99d)=7$. The difference is always $7$, the difference of the first terms: the $d$'s cancel.
6 $7(a+6d)=11(a+10d)$ gives $7a+42d=11a+110d$, so $4a=-68d$ and $a=-17d$. Then $a_{18}=a+17d=-17d+17d=0$.
1 Find the $11$th term from the end of the AP $10, 7, 4, \dots, -62$.
2 Find the $5$th term from the end of the AP $17, 14, 11, \dots, -40$.
3 Find the middle term of the AP $213, 205, 197, \dots, 37$.
4Stadium. The front row of a stand has $20$ seats and each row has $4$ more seats than the row in front. How many seats are in the $25$th row? Which row has $64$ seats?
5Taxi fare. A taxi charges ₹ $40$ for the first kilometre and ₹ $18$ for each further kilometre. Find the fare for a $15$ km ride, and the length of a ride that costs ₹ $382$.
6Car value. A car bought for ₹ $6\,50\,000$ loses ₹ $45\,000$ of value every year. Find its value at the end of the $8$th year. In which year does its value first fall below ₹ $2\,00\,000$?
7Irrigation tank. At $6$ a.m. a tank holds $2400$ L, and $150$ L flow out to the fields every hour. How much is left at $1$ p.m., and at what time is the tank empty?
💡 Show solutions▶
1 $l=-62$, $d=-3$: $\;l-10d=-62-10(-3)=-32$. (Check: there are $25$ terms, and the $15$th is $10+14(-3)=-32$ ✓.)
2 $l=-40$, $d=-3$: $\;-40-4(-3)=-28$.
3 $213+(m-1)(-8)=37$ gives $m-1=22$, $m=23$ terms. The middle term is the $12$th: $\;213+11(-8)=125$.
4 $a_{25}=20+24\times4=116$ seats. $\;20+(n-1)\times4=64$ gives $n-1=11$: the $12$th row.
5 Fares form the AP $40, 58, 76, \dots$ (₹), with the $n$th term the fare for $n$ km. $\;a_{15}=40+14\times18=$ ₹ $292$. $\;40+(n-1)\times18=382$ gives $n-1=19$: a $20$ km ride.
6 Values at the end of years $1, 2, 3, \dots$: ₹ $6\,05\,000, 5\,60\,000, \dots$, so $a=605\,000$, $d=-45\,000$. $\;a_8=605\,000-7\times45\,000=$ ₹ $2\,90\,000$. Below ₹ $2\,00\,000$: $605\,000-45\,000(n-1)\lt200\,000$ gives $n-1\gt9$, so $n=11$ (at the end of year $10$ it is exactly ₹ $2\,00\,000$; during year $11$ it falls below).
7 Amounts at $6, 7, 8, \dots$ a.m.: $2400, 2250, 2100, \dots$ L. $1$ p.m. is $7$ hours later, the $8$th term: $\;2400+7(-150)=1350$ L. Empty: $2400-150\times h=0$ gives $h=16$ hours after $6$ a.m., i.e. at $10$ p.m.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main
How this is examined: the $n$th term appears in almost every board paper — finding a term, "which term is …", the first negative term, an AP from two terms, and the $n$th term from the end, often inside a case study about seats, savings or fares. Always write $a_n=a+(n-1)d$ with your values substituted; and when you find $n$, say whether it is a whole number.
E1.Board · 1 mark The $21$st term of the AP whose first two terms are $-3$ and $4$ is (A) $17$ (B) $137$ (C) $143$ (D) $-143$
💡 Show solution▶
Idea $a=-3$, $d=4-(-3)=7$, so $a_{21}=-3+20\times7=137$.
Traps (C) comes from taking $a=+3$; (A) comes from $d=4-3=1$ with $a=-3$. Always find $d$ as second term minus first term, signs included.
Answer (B) $137$.
E2.Board · 1 mark The $n$th term of an AP is $a_n=5-3n$. Its common difference is (A) $5$ (B) $3$ (C) $-3$ (D) $2$
💡 Show solution▶
Idea $a_1=2$, $a_2=-1$, so $d=-1-2=-3$. (In general, $a_{n+1}-a_n=-3$: the coefficient of $n$.)
Trap (D) is the first term, not the common difference; (A) is the constant in the formula.
Answer (C) $-3$.
E3.Board · 2 marks Which term of the AP $65, 61, 57, 53, \dots$ is its first negative term?
💡 Show solution▶
Inequality $a=65$, $d=-4$. We need $65+(n-1)(-4)\lt0$, i.e. $4(n-1)\gt65$, $n-1\gt16.25$. (1 mark)
Answer The smallest whole $n$ is $18$: $\;a_{18}=65-68=-3$, while $a_{17}=65-64=1$ is still positive. The $18$th term. (1 mark)
E4.Board · 3 marks The $5$th term of an AP is $26$ and its $10$th term is $51$. Find the AP and its $25$th term. Is $300$ a term of this AP?
💡 Show solution▶
AP $a+4d=26$ and $a+9d=51$; subtracting, $5d=25$, $d=5$, $a=6$. The AP is $6, 11, 16, 21, \dots$ (1 mark)
25th term $a_{25}=6+24\times5=126$. (1 mark)
Is 300 a term? $6+(n-1)\times5=300$ gives $n-1=\tfrac{294}5=58.8$, not a whole number. So $300$ is not a term. (1 mark)
E5.Board · Case study A school auditorium has $30$ rows of seats. The first row has $18$ seats, and each row has $2$ seats more than the row in front of it. (i) How many seats are there in the $15$th row? (1 mark) (ii) How many seats are in the last row, and in the $6$th row from the back? (1 mark) (iii) Which row has $50$ seats? Show your working. (2 marks)
💡 Show solution▶
The AP Seats per row: $18, 20, 22, \dots$ with $a=18$, $d=2$ and $30$ terms.
(i) $a_{15}=18+14\times2=46$ seats.
(ii) Last row: $l=a_{30}=18+29\times2=76$ seats. $6$th from the back: $l-5d=76-10=66$ seats (it is the $25$th row: $18+24\times2=66$ ✓).
(iii) $18+(n-1)\times2=50$ gives $(n-1)\times2=32$, $n-1=16$, $n=17$. The $17$th row has $50$ seats.
E6.JEE Main In an AP, the $p$th term is $q$ and the $q$th term is $p$, where $p\ne q$. Then its $(p+q)$th term is (A) $0$ (B) $p+q$ (C) $p-q$ (D) $1$
💡 Show solution▶
Equations $a+(p-1)d=q$ and $a+(q-1)d=p$. Subtracting, $(p-q)d=q-p$, so $d=-1$ (we may divide by $p-q\ne0$).
First term $a=q-(p-1)(-1)=p+q-1$.
Answer $a_{p+q}=(p+q-1)+(p+q-1)(-1)=0$: (A). Check with numbers: $p=2$, $q=5$ gives the AP $6, 5, 4, 3, 2, 1, 0, \dots$ — the $2$nd term is $5$, the $5$th term is $2$, and the $7$th term is $0$ ✓. (B) is the value of $a+1$, a tempting slip.
🔬 Research–Teaching NexusMini project · optional, about 40 minutes
Research question. How many steps does it take to climb one floor of your school or apartment building — and can the $n$th-term formula predict the height of any step, or the number of steps to the top floor?
Why it matters. Architects follow building codes that fix the height of each step (the "riser", often $15$–$18$ cm) so that every step is the same — a fall on an uneven staircase is far more likely. Equal risers mean the heights of the step tops form an AP, so a designer can compute how many steps a floor needs before a single brick is laid.
What to do.
Choose one flight of stairs. Measure the height of $5$ different risers with a tape measure (in cm). Are they equal? Take the average as $d$.
The top of step $n$ is $a_n=a+(n-1)d$ cm above the floor, where $a$ is the height of the first step. Write this formula for your staircase.
Measure the total height of one floor (from one floor level to the next) and use the formula to predict how many steps it needs. Count the steps and compare.
Predict the height above the ground floor of the $n$th step counted from the bottom of the building (continue the AP across floors, ignoring landings). Which step number would be at $10$ m?
Find the $5$th step from the top of your flight using $l-(n-1)d$, and check it by measuring.
Tools: Tape measure, notebook or phone spreadsheet
Hand in: Your riser measurements and average $d$, your formula for $a_n$, the predicted and actual number of steps per floor, the step number at $10$ m, and a short note on how uneven the risers were.
Going further. Find out the riser height recommended by India's National Building Code (or your country's code). Does your staircase meet it? How would the number of steps change if the riser were $2$ cm shorter?
Teacher/parent note: with risers of $16.5$ cm and a floor-to-floor height of $3.3$ m, a floor needs $330\div16.5=20$ risers. The step number at $10$ m is the smallest $n$ with $n\times16.5\ge1000$, i.e. $n=61$ (when $a=d$). Students should see that a real AP is only approximately an AP because of measurement error.
🧪 Self-check quiz
0/5 correct
Q1. The $10$th term of $2, 5, 8, \dots$ is
Q2. Which term of $4, 9, 14, \dots$ is $89$?
Q3. The $7$th term from the end of the AP $1, 4, 7, \dots, 52$ is
Q4. In an AP, $a_4=9$ and $a_6=15$. The first term is