nth Term of an AP: Formula and Worked Examples, Class 10

The result#

The $n$th term of the AP $a, a + d, a + 2d, \ldots$ is

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

For a finite AP whose last term is the $n$th term, $l = a + (n - 1)d$.

Why the for­mula is true#

Each step from one term to the next adds one $d$. Write the first few terms and count the $d$s:

$$\begin{aligned} a_1 &= a = a + 0 \cdot d \\ a_2 &= a_1 + d = a + 1 \cdot d \\ a_3 &= a_2 + d = a + 2d \\ a_4 &= a_3 + d = a + 3d \end{aligned}$$

To travel from the 1st term to the $n$th term you take $n - 1$ steps, because posi­tions 1 to $n$ have $n - 1$ gaps between them. Each step adds $d$, so alto­gether $(n - 1)d$ has been added to $a$. There­fore $a_n = a + (n - 1)d$.

A use­ful con­se­quence: for any two posi­tions $p$ and $q$, $a_p - a_q = (p - q)d$. The gap between two terms is the num­ber of steps between them times $d$.

The graph shows another way to see the for­mula. For 5, 9, 13, …, $a_n = 5 + (n - 1) \times 4 = 4n + 1$. The $n$th term of any AP is a lin­ear expres­sion in $n$, and the coef­fi­cient of $n$ is the com­mon dif­fer­ence.

Method: using the for­mula#

1. Read off $a$ (the first term) and find $d$ (sec­ond term minus first term).
2. Write the for­mula $a_n = a + (n - 1)d$ and decide which quan­tity is unknown: $a_n$, $n$, $a$ or $d$.
3. Sub­sti­tute the known val­ues and solve the equa­tion for the unknown.
4. If you solved for $n$, check that it is a pos­i­tive whole num­ber. If it is not, the given value is not a term of the AP.
5. Check your answer by sub­sti­tut­ing back, and state it in words.

Worked exam­ples#

Exam­ple 1: find­ing a term#

Find the 20th term of the AP 5, 9, 13, … .

Here $a = 5$ and $d = 9 - 5 = 4$. With $n = 20$:

$$a_{20} = 5 + (20 - 1) \times 4 = 5 + 76 = 81$$

Answer: The 20th term is 81.

Exam­ple 2: which term has a given value#

Which term of the AP 7, 13, 19, … is 205?

Here $a = 7$ and $d = 6$. Put $a_n = 205$:

$$\begin{aligned} 7 + (n - 1) \times 6 &= 205 \\ (n - 1) \times 6 &= 198 \\ n - 1 &= 33 \\ n &= 34 \end{aligned}$$

Answer: 205 is the 34th term.

Exam­ple 3: is a num­ber a term?#

Is 150 a term of the AP 11, 17, 23, … ?

Here $a = 11$ and $d = 6$. Sup­pose $a_n = 150$:

$$\displaystyle \begin{aligned} 11 + (n - 1) \times 6 &= 150 \\ (n - 1) \times 6 &= 139 \\ n - 1 &= \frac{139}{6} \end{aligned}$$

Since $\displaystyle \frac{139}{6}$ is not a whole num­ber, $n$ is not a whole num­ber, and a term posi­tion must be a pos­i­tive whole num­ber.

Answer: No, 150 is not a term of this AP.

Why this works: every term is 11 more than a mul­ti­ple of 6, and $150 - 11 = 139$ is not a mul­ti­ple of 6.

Exam­ple 4: find­ing the AP from two terms#

The 4th term of an AP is 17 and its 11th term is 45. Find the AP.

Write both facts using the for­mula:

$$a + 3d = 17 \qquad \text{and} \qquad a + 10d = 45$$

Sub­tract the first equa­tion from the sec­ond:

$$7d = 28 \quad \Rightarrow \quad d = 4$$

Sub­sti­tute into $a + 3d = 17$: $a + 12 = 17$, so $a = 5$.

Check: $a_{11} = 5 + 10 \times 4 = 45$.

Answer: The AP is 5, 9, 13, 17, … .

Exam­ple 5: a term counted from the end#

Find the 8th term from the end of the AP 4, 9, 14, …, 254.

Count­ing back­wards from the last term is an AP too, with first term $l$ and com­mon dif­fer­ence $-d$. So the $m$th term from the end is

$$l - (m - 1)d$$

Here $l = 254$, $d = 5$ and $m = 8$:

$$254 - (8 - 1) \times 5 = 254 - 35 = 219$$

Check by count­ing from the front. The num­ber of terms is found from $4 + (n - 1) \times 5 = 254$, giv­ing $n - 1 = 50$, so $n = 51$. The 8th term from the end is the $(51 - 8 + 1)$th $=$ 44th term from the start, and $a_{44} = 4 + 43 \times 5 = 219$.

Answer: 219.

Exam­ple 6: count­ing mul­ti­ples in a range#

How many three-digit num­bers are divis­i­ble by 9?

The small­est three-digit mul­ti­ple of 9 is 108 (since $9 \times 11 = 99$ has only two dig­its and $9 \times 12 = 108$), and the largest is 999 (since $9 \times 111 = 999$). The mul­ti­ples form the AP 108, 117, …, 999 with $d = 9$.

$$\begin{aligned} 108 + (n - 1) \times 9 &= 999 \\ (n - 1) \times 9 &= 891 \\ n - 1 &= 99 \\ n &= 100 \end{aligned}$$

Answer: 100 three-digit num­bers are divis­i­ble by 9.

Exam­ple 7: a sav­ings word prob­lem#

Meera saves ₹150 in the first month and increases her sav­ing by ₹25 every month after that. How much does she save in the 12th month, and in which month does she first save ₹600?

The monthly sav­ings form an AP with $a = 150$ and $d = 25$.

For the 12th month:

$$a_{12} = 150 + 11 \times 25 = 150 + 275 = 425$$

For a sav­ing of ₹600:

$$\begin{aligned} 150 + (n - 1) \times 25 &= 600 \\ (n - 1) \times 25 &= 450 \\ n - 1 &= 18 \\ n &= 19 \end{aligned}$$

Answer: She saves ₹425 in the 12th month, and she saves ₹600 in the 19th month.

Com­mon mis­takes#

• Writ­ing $a_n = a + nd$. The first term already sits in posi­tion 1, so only $n - 1$ steps are needed.
• Tak­ing $d$ as first term minus sec­ond term. For 26, 22, 18 the com­mon dif­fer­ence is $-4$, not 4.
• Accept­ing a frac­tional or neg­a­tive value of $n$. A term num­ber must be a pos­i­tive whole num­ber; oth­er­wise the value is not a term.
• Using $l + (m - 1)d$ for the $m$th term from the end. Going back­wards sub­tracts $d$.
• Count­ing mul­ti­ples by divid­ing the range, such as $\displaystyle \frac{999 - 100}{9}$. Find the first and last mul­ti­ples and use the for­mula instead.
• Stop­ping at $d$ when the ques­tion asks for the AP. Always find $a$ too and write out the first few terms.

Try these#

1. Find the 15th term of the AP 3, $-1$, $-5$, … . Answer: $-53$.
2. Which term of the AP 26, 22, 18, … is $-82$? Answer: The 28th term.
3. How many two-digit num­bers are divis­i­ble by 4? Answer: 22.
4. Find the 10th term from the end of the AP 7, 10, 13, …, 184. Answer: 157.
5. The 3rd term of an AP is 12 and the 9th term is 30. Find its 20th term. Answer: 63 (the AP is 6, 9, 12, …).

Key terms#

nth term (gen­eral term)

The term in posi­tion $n$, given by $a_n = a + (n - 1)d$.

First term

The term in posi­tion 1, writ­ten $a$.

Com­mon dif­fer­ence

The con­stant $d = a_{k+1} - a_k$ between con­sec­u­tive terms.

Last term

The final term $l$ of a finite AP; $l = a + (n - 1)d$.

Num­ber of terms

The value of $n$ for which $a_n = l$; it must be a pos­i­tive whole num­ber.

Term from the end

The $m$th term count­ing back­wards from $l$, equal to $l - (m - 1)d$.

Com­mon ques­tions#

Why is it n minus 1 and not n?#

Between posi­tion 1 and posi­tion $n$ there are $n - 1$ gaps. Only the gaps add a $d$, so the $n$th term is $a$ plus $n - 1$ lots of $d$.

How can I tell quickly that a num­ber is not a term?#

Sub­tract the first term from the num­ber. If the result is not an exact mul­ti­ple of $d$ (with the right sign), the num­ber is not a term.

Is the nth term always a lin­ear expres­sion?#

Yes. $a + (n - 1)d = dn + (a - d)$, which is of the form $pn + q$ with $p = d$. Con­versely, any sequence with $a_n = pn + q$ is an AP with com­mon dif­fer­ence $p$.

How do I find the mid­dle term of a finite AP?#

If the num­ber of terms $n$ is odd, the mid­dle term is the $\displaystyle \frac{n + 1}{2}$th term. If $n$ is even, there are two mid­dle terms, the $\displaystyle \frac{n}{2}$th and the $\displaystyle \left(\frac{n}{2} + 1\right)$th.

Can two dif­fer­ent terms of an AP be equal?#

Only when $d = 0$. If $d \ne 0$, then $a_p - a_q = (p - q)d$ is non-zero when­ever $p \ne q$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.