Sum of First n Terms of an AP: Class 10 Formula and Examples

The sum for­mula#

The result#

The sum of the first $n$ terms of an AP with first term $a$ and com­mon dif­fer­ence $d$ is

$$\displaystyle S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$$

If the last of these $n$ terms is $l = a + (n - 1)d$, this can also be writ­ten as

$$\displaystyle S_n = \frac{n}{2}(a + l)$$

Proof#

Write the sum for­wards, and then back­wards start­ing from the $n$th term $a + (n - 1)d$:

$$\begin{aligned} S_n &= a + (a + d) + (a + 2d) + \cdots + [a + (n - 1)d] \\ S_n &= [a + (n - 1)d] + [a + (n - 2)d] + \cdots + a \end{aligned}$$

Add the two lines term by term. In each col­umn, one term has $k$ lots of $d$ and the term beneath it has $n - 1 - k$ lots, so each col­umn adds up to

$$2a + (n - 1)d$$

There are $n$ columns, so

$$\displaystyle 2S_n = n\left[2a + (n - 1)d\right] \quad \Rightarrow \quad S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$$

Since $2a + (n - 1)d = a + [a + (n - 1)d] = a + l$, this is the same as $\displaystyle S_n = \frac{n}{2}(a + l)$. In words: the sum is the num­ber of terms times the aver­age of the first and last terms.

Sum of the first n nat­ural num­bers#

The nat­ural num­bers $1, 2, 3, \ldots, n$ form an AP with $a = 1$ and $l = n$. So

$$\displaystyle 1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}$$

For exam­ple, $\displaystyle 1 + 2 + \cdots + 40 = \frac{40 \times 41}{2} = 820$.

Get­ting a term back from the sums#

The sum of the first $n$ terms is the sum of the first $n - 1$ terms plus the $n$th term: $S_n = S_{n-1} + a_n$. There­fore, for $n \ge 2$,

$$a_n = S_n - S_{n-1}$$

and $a_1 = S_1$. This lets you find an AP when you are only told a for­mula for its sums.

Method#

1. Iden­tify $a$ and $d$, and decide whether you know $n$, the last term $l$, or the sum $S_n$.
2. If you know $l$ but not $n$, first find $n$ from $l = a + (n - 1)d$.
3. Use $\displaystyle S_n = \frac{n}{2}(a + l)$ when the last term is known; oth­er­wise use $\displaystyle S_n = \frac{n}{2}[2a + (n - 1)d]$.
4. If $n$ is unknown and $S_n$ is given, sub­sti­tute, clear the frac­tion and solve the qua­dratic in $n$.
5. Keep only pos­i­tive whole-num­ber val­ues of $n$. If two such val­ues appear, check both and explain why both work.

Worked exam­ples#

Exam­ple 1: sum of a given num­ber of terms#

Find the sum of the first 20 terms of the AP 3, 8, 13, … .

Here $a = 3$, $d = 5$ and $n = 20$:

$$\displaystyle S_{20} = \frac{20}{2}\left[2 \times 3 + 19 \times 5\right] = 10\,[6 + 95] = 10 \times 101 = 1010$$

Answer: 1010.

Exam­ple 2: sum when the last term is given#

Find $5 + 9 + 13 + \cdots + 101$.

First find the num­ber of terms, with $a = 5$, $d = 4$ and $l = 101$:

$$5 + (n - 1) \times 4 = 101 \;\Rightarrow\; n - 1 = 24 \;\Rightarrow\; n = 25$$

Now use the first-and-last form:

$$\displaystyle S_{25} = \frac{25}{2}(5 + 101) = \frac{25}{2} \times 106 = 25 \times 53 = 1325$$

Answer: 1325.

Exam­ple 3: sum of odd num­bers#

Find the sum of all odd num­bers from 1 to 99.

The odd num­bers 1, 3, 5, …, 99 form an AP with $a = 1$, $d = 2$ and $l = 99$. From $1 + (n - 1) \times 2 = 99$ we get $n = 50$. So

$$\displaystyle S_{50} = \frac{50}{2}(1 + 99) = 25 \times 100 = 2500$$

Answer: 2500. (Notice that $2500 = 50^2$: the sum of the first $n$ odd num­bers is $n^2$.)

Exam­ple 4: find­ing n and reject­ing an invalid root#

How many terms of the AP 3, 7, 11, … must be taken so that their sum is 406?

Here $a = 3$ and $d = 4$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[6 + (n - 1) \times 4\right] &= 406 \\ \frac{n}{2}(4n + 2) &= 406 \\ n(2n + 1) &= 406 \\ 2n^2 + n - 406 &= 0 \end{aligned}$$

The dis­crim­i­nant is $1 + 4 \times 2 \times 406 = 3249 = 57^2$, so

$$\displaystyle n = \frac{-1 \pm 57}{4} \quad \Rightarrow \quad n = 14 \quad \text{or} \quad n = -\frac{29}{2}$$

The num­ber of terms must be a pos­i­tive whole num­ber, so $\displaystyle n = -\frac{29}{2}$ is rejected.

Answer: 14 terms.

Exam­ple 5: when two val­ues of n both work#

How many terms of the AP 30, 27, 24, … must be taken to give a sum of 135?

Here $a = 30$ and $d = -3$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[60 - 3(n - 1)\right] &= 135 \\ n(63 - 3n) &= 270 \\ 3n^2 - 63n + 270 &= 0 \\ n^2 - 21n + 90 &= 0 \\ (n - 6)(n - 15) &= 0 \end{aligned}$$

So $n = 6$ or $n = 15$. Both are pos­i­tive whole num­bers, so both are valid. Why? The $n$th term is $a_n = 30 - 3(n - 1) = 33 - 3n$, so the terms from the 7th to the 15th are

$$12, 9, 6, 3, 0, -3, -6, -9, -12$$

These nine terms add up to 0, so adding them to the first six terms does not change the total.

Answer: Either 6 terms or 15 terms give a sum of 135.

Exam­ple 6: find­ing the AP from its sums#

The sum of the first $n$ terms of an AP is $S_n = 2n^2 + 5n$. Find the $n$th term and the com­mon dif­fer­ence.

Use $a_n = S_n - S_{n-1}$:

$$\begin{aligned} a_n &= (2n^2 + 5n) - \left[2(n - 1)^2 + 5(n - 1)\right] \\ &= 2n^2 + 5n - (2n^2 - 4n + 2 + 5n - 5) \\ &= 4n + 3 \end{aligned}$$

Check with $n = 1$: $a_1 = 7$ and $S_1 = 2 + 5 = 7$. The terms are 7, 11, 15, …, so $d = 4$.

Answer: $a_n = 4n + 3$ and $d = 4$.

Exam­ple 7: word prob­lems#

(a) Pipes are stacked with 24 pipes in the bot­tom row, 23 in the next, 22 in the next, and so on, with 9 pipes in the top row. How many pipes are there?

The rows form the AP 24, 23, …, 9 with $d = -1$. From $24 - (n - 1) = 9$ we get $n = 16$ rows. So the total is

$$\displaystyle S_{16} = \frac{16}{2}(24 + 9) = 8 \times 33 = 264$$

(b) A loan is repaid in monthly instal­ments of ₹1000, ₹1100, ₹1200, and so on. How much is paid in the first 12 months?

$$\displaystyle S_{12} = \frac{12}{2}\left[2 \times 1000 + 11 \times 100\right] = 6 \times 3100 = 18600$$

Answer: (a) 264 pipes; (b) ₹18,600.

Com­mon mis­takes#

• Using $\displaystyle S_n = \frac{n}{2}(a + l)$ with the wrong $n$. Find the num­ber of terms from $l = a + (n - 1)d$ first; do not guess it.
• For­get­ting the brack­ets, for exam­ple writ­ing $\displaystyle \frac{n}{2} \times 2a + (n - 1)d$. The $\displaystyle \frac{n}{2}$ mul­ti­plies the whole bracket.
• Keep­ing a neg­a­tive or frac­tional value of $n$. A num­ber of terms must be a pos­i­tive whole num­ber.
• Throw­ing away a valid sec­ond root. When both roots are pos­i­tive whole num­bers, both answers are cor­rect; explain why.
• Con­fus­ing $a_n$ with $S_n$. The $n$th term is one num­ber; $S_n$ is the total of the first $n$ num­bers.
• Writ­ing $a_n = S_n - S_{n-1}$ for $n = 1$. For the first term use $a_1 = S_1$.

Try these#

1. Find the sum of the first 15 terms of the AP 8, 5, 2, … . Answer: $-195$.
2. Find the sum of all mul­ti­ples of 6 between 1 and 200. Answer: 3366.
3. If $S_n = n^2 + 4n$, find $a_n$ and $a_{10}$. Answer: $a_n = 2n + 3$; $a_{10} = 23$.
4. How many terms of 45, 42, 39, … give a sum of 315? Answer: 10 or 21 (the 11th to 21st terms add up to 0).
5. Find $1 + 2 + 3 + \cdots + 60$. Answer: 1830.

Key terms#

Sum of n terms

$S_n$, the total of the first $n$ terms of an AP.

Sum for­mula

$\displaystyle S_n = \frac{n}{2}[2a + (n - 1)d]$.

First-and-last form

$\displaystyle S_n = \frac{n}{2}(a + l)$, used when the last term $l$ is known.

Pair­ing method

Writ­ing a sum for­wards and back­wards so that every col­umn has the same total.

Nat­ural num­ber sum

$\displaystyle 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$.

Term from sums

$a_n = S_n - S_{n-1}$ for $n \ge 2$, with $a_1 = S_1$.

Com­mon ques­tions#

Which sum for­mula should I use?#

If the last term is known, $\displaystyle \frac{n}{2}(a + l)$ is quicker. If you know $d$ and $n$ but not the last term, use $\displaystyle \frac{n}{2}[2a + (n - 1)d]$. Both always give the same answer.

Can the sum of an AP be zero or neg­a­tive?#

Yes. When $d$ is neg­a­tive the later terms become neg­a­tive, and the total can fall to zero or below, as in the AP 12, 9, …, $-12$, whose terms add up to 0.

Why can two dif­fer­ent val­ues of n give the same sum?#

If a block of con­sec­u­tive terms adds up to zero, includ­ing or leav­ing out that block does not change the total. This can hap­pen only when the terms change sign, so $d$ and $a$ have oppo­site signs.

Does the pair­ing method need an even num­ber of terms?#

No. Writ­ing the sum twice avoids the prob­lem of a mid­dle term, because every col­umn pairs a term with a dif­fer­ent copy. That is why the for­mula works for odd $n$ too.

How can I check an answer quickly?#

For small $n$, add the terms directly. For large $n$, check that $S_1$ equals the first term and that $S_2 - S_1$ equals the sec­ond term.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.
3. Gelfand, I. M. and Shen, A. Alge­bra. Birkhäuser.