# Sum of n Terms of an AP

Sometime in the 19th century, somewhere in Germany, a Math class was going on. The students were all 10-year olds or so. The teacher, to keep the students occupied for a while, and to maintain silence, asked his students to sum all the numbers from 1 up to 100. The teacher had barely assigned this problem and students had barely started using their chalk and slate when one particular boy shouted out the answer: *Its 5050*. Legend has it that the teacher was so stunned that he kept staring at the boy continuously for about an entire hour afterwards! This boy was the great German mathematician Carl Friedrich Gauss.

How did he arrive at the sum so quickly? Well, he noticed that terms equidistant from the beginning and end of the series had a constant sum equal to 101:

\[\begin{array}{l}1 + 100 = 101\\2 + 99 = 101\\\,\,\,\,\,\,\,\,\,\, \vdots \\50 + 51 = 101\end{array}\]

Now, there are 50 such pairs possible, so the sum is simply 50 × 101 = 5050. Neat!

Can we apply this logic to other APs? Consider the following AP:

\[2,\,\,7,\,\,12,\,\,17,\,...,102\]

You can verify that there are 21 terms in this AP. How can we evaluate the sum of these 21 terms, without summing them explicitly? Once again, note that terms equidistant from the beginning and end of the series have a constant sum. Let’s use this fact as follows. Write the sum *s* of these terms twice, once with the terms in the order given, and once with the terms in the reverse order, as shown:

\[\begin{array}{l}S = 2 + 7 + 12 + \,... + 92 + 97 + 102\\S = 102 + 97 + 92 + ...12 + 7 + 2\end{array}\]

Now, let us sum the two expressions (on the right side, we sum the corresponding terms):

\[\begin{align}2S = &\left( \begin{array}{l}\,\,\,2\\\,\, + \\102\end{array} \right) + \left( \begin{array}{l}\,7\\ + \\97\end{array} \right) + \left( \begin{array}{l}12\\\, + \\92\end{array} \right) + ...\left( \begin{array}{l}92\\\, + \\12\end{array} \right) + \left( \begin{array}{l}97\\\, + \\\,7\end{array} \right) + \left( \begin{array}{l}102\\\,\, + \\\,\,2\end{array} \right)\\& = \underbrace {104 + 104 + 104 + ... + 104 + 104 + 104}_{{\rm{21 times}}}\\& = 104 \times 21\\&= 2184\end{align}\]

Clearly, the observation Gauss made helps us in summing any number of terms of an AP without having to calculate their sum explicitly by adding them one by one.

Taking cue from the previous example, let us develop a general formula to enable us to calculate the sum of an arbitrary number of terms of any AP.

Consider an AP with first term equal to *a* and common difference equal to *d*. Let *s* be the sum of the first *n* terms of this AP. Our task is to develop a formula for *s*. We follow exactly the same approach as earlier. We write out the series for *s* twice, once in the original order, and once it the reverse order, and sum the corresponding terms. Each pair of such terms will turn out to have the same sum:

\[\begin{array}{l}\left\{ \begin{array}{l}S = a + \left( {a + d} \right) + ... + \left( {a + \left( {n - 2} \right)d} \right) + \left( {a + \left( {n - 1} \right)d} \right)\\S = \left( {a + \left( {n - 1} \right)d} \right) + \left( {a + \left( {n - 2} \right)d} \right) + ... + \left( {a + d} \right) + a\end{array} \right.\\ \Rightarrow \,\,\,2S = \underbrace {\left( {2a + \left( {n - 1} \right)d} \right) + \left( {2a + \left( {n - 1} \right)d} \right) + ...}_{n{\rm{ times}}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = n\left( {2a + \left( {n - 1} \right)d} \right)\\ \Rightarrow \,\,\,S = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\end{array}\]

This is the general formula for the sum of the first *n* terms of any AP. We can express it in an alternate form as follows:

\[\begin{align}&S = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\\&= \frac{n}{2}\left( {\underbrace a_{{\rm{First \,\, term}}} + \underbrace {a + \left( {n - 1} \right)d}_{{\rm{Last\,\, term,\,\, that \,\, is, }}{n^{th}}{\rm{ term}}}} \right)\\&= \frac{n}{2}\left( {a + l} \right)\end{align}\]

This formula expresses the sum of the first *n* terms in terms of the first term and the *n*-th term (the *last* term of the sum).

Let us apply these formulas on an example. Consider the following sum of the terms of an AP:

\[S = - 1 + 5 + 11 + 17 + ... + 149\]

It can be verified that there are 26 terms in this AP. Thus:

\[a = - 1,\,\,\,d = 6,\,\,\,n = 26,\,\,\,l = 149\]

Using the first formula, we have:

\[\begin{align}&S = \frac{{26}}{2}\left( {2\left( { - 1} \right) + \left( {26 - 1} \right)6} \right)\\& = 13\left( { - 2 + 150} \right)\\&= 13\left( {148} \right)\\&= 1924\end{align}\]

Using the second formula, we have:

\[\begin{align}&S = \frac{{26}}{2}\left( {\underbrace { - 1}_{{\rm{First \,\,term}}} + \underbrace {149}_{{\rm{Last \,\,term}}}} \right)\\&= 13\left( {148} \right)\\& = 1924\end{align}\]

** Remark:** The sum of the first

*n*terms of a progression is typically represented as \({S_n}\).

**Example 1:** What is the sum of the first *n* positive integers?

**Solution:** We have:

\[\begin{align}&S = 1 + 2 + 3 + ... + n\\& = \frac{n}{2}\left( {\underbrace {\,\,\,1\,\,\,}_{{\rm{First \,\, term}}} + \underbrace {\,\,\,n\,\,\,}_{{\rm{Last \,\, term}}}} \right)\\& = \frac{{n\left( {n + 1} \right)}}{2}\end{align}\]

**Example 2:** Evaluate the following sum:

\[S = \underbrace {190 + 167 + 144 + 121 + ...}_{20\,\,{\rm{ terms}}}\]

**Solution:** We do not know the last term in this sequence, so we will use the first formula to calculate this sum. We note that:

\[a = 190,\,\,\,d = - 23,\,\,\,n = 20\]

Thus,

\[\begin{align}&S = \frac{{20}}{2}\left( {2\left( {190} \right) + \left( {20 - 1} \right)\left( { - 23} \right)} \right)\\&\;\;= 10\left( {380 - 437} \right)\\& \;\;= 10\left( { - 57} \right)\\&\;\; = - 570\end{align}\]

**Example 3:** Consider the following AP:

\[24,\,\,21,\,\,18,\,...\]

How many terms of this AP must be taken so that their sum is 78?

**Solution:** Let the number of terms be *n*. We have:

\[a = 24,\,\,\,d = - 3,\,\,\,S = 78\]

Now,

\[\begin{align}&S = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\\&\Rightarrow \,\,\,78 = \frac{n}{2}\left( {48 - 3\left( {n - 1} \right)} \right)\\&\,\;\;\;\;\;\;\;\;\; = \frac{n}{2}\left( {51 - 3n} \right)\\& \Rightarrow \,\,\,3{n^2} - 51n + 156 = 0\\& \Rightarrow \,\,\,{n^2} - 17n + 52 = 0\\ &\Rightarrow \,\,\,\left( {n - 4} \right)\left( {n - 13} \right) = 0\\& \Rightarrow \,\,\,n = 4,\,\,13\end{align}\]

You might be surprised that we have obtained two values of *n*. However, a little thinking will shed light on the reason for this. The sum of the first 4 terms of this AP is 78. As you take more and more terms into the sum, the sum will increase but only up to a certain point, because after a point, the terms in the AP become negative. So, any more terms will decrease the sum. This way, the sum of the first 13 terms also comes out to 78.

As an exercise, find the maximum value that the sum of the first certain number of terms of this AP can take.

**Example 4:** The sum of the first *p* terms of an AP is equal to the sum of the first *q* terms. What is the sum of the first *p* + *q* terms?

**Solution:** Let the first term of the AP be *a*, and its common difference be *d*. We have:

\[\begin{align}&{S_p} = {S_q}\\& \Rightarrow \,\,\,\frac{p}{2}\left( {2a + \left( {p - 1} \right)d} \right) = \frac{q}{2}\left( {2a + \left( {q - 1} \right)d} \right)\\ &\Rightarrow \,\,\,2ap + p\left( {p - 1} \right)d = 2aq + q\left( {q - 1} \right)d\\ &\Rightarrow \,\,\,2a\left( {p - q} \right) + \left( {\left( {{p^2} - {q^2}} \right) - \left( {p - q} \right)} \right)d = 0\,\,\,\,\,{\rm{(how?)}}\\& \Rightarrow \,\,\,2a + \left( {p + q - 1} \right)d = 0\,\,\,\,\,{\rm{(how?)}}\end{align}\]

Make sure that you understand each step. Now, let us evaluate the sum of the first *p* + *q* terms:

\[\begin{array}{l}{S_{p + q}} = \frac{{\left( {p + q} \right)}}{2}\left( {2a + \left( {p + q - 1} \right)d} \right)\\\;\;\;\;\;\,\, = \frac{{\left( {p + q} \right)}}{2} \times 0\\\;\;\;\;\;\,\, = 0\end{array}\]

Thus, the sum of the first *p* + *q* terms is 0. A very interesting result indeed!

**Example 5:** The sum of *n* terms of an AP is \(3{n^2} + 5n\). The *m*^{th} term of the AP is 164. Find the value of *m*.

**Solution:** We will find an expression for the *n*^{th} term. To do that, we will find the difference between the sum of *n* terms and the sum of \(\left( {n - 1} \right)\) terms:

\[\begin{align}{T_n}& = {S_n} - {S_{n - 1}}\\&= \left( {3{n^2} + 5n} \right) - \left\{ {3{{\left( {n - 1} \right)}^2} + 5\left( {n - 1} \right)} \right\}\\&= 3\left\{ {{n^2} - {{\left( {n - 1} \right)}^2}} \right\} + 5\left\{ {n - \left( {n - 1} \right)} \right\}\\&= 3\left( {2n - 1} \right) + 5 = 6n + 2\end{align}\]

Since the *m*^{th} term is 164, we have:

\[\begin{array}{l}{T_m} = 6m + 2 = 164\\ \Rightarrow \,\,\,m = 27\end{array}\]

**Example 6:** The first term of an AP is 11. The sum of its first four terms is 56, and its last four terms is 112. Find the number of terms in the AP.

**Solution:** Let there be *n* terms in the AP, let *d *be the common difference, and let the last term be *l*. We have:

\[l = 11 + \left( {n - 1} \right)d\]

Now, the sum of the first four terms will be:

\[\begin{align}&{S_{{\rm{first\,\, four}}}} = \frac{4}{2}\left( {22 + 3d} \right) = 56\\& \Rightarrow \,\,\,d = 2\end{align}\]

And the sum of the last four terms will be:

\[\begin{align}{S_{{\rm{last \,\,four}}}} &= \frac{4}{2}\left( {2l - 3d} \right) = 112\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\Rightarrow \,\,\,2l - 3d = 56\\& \Rightarrow \,\,\,l = 31\end{align}\]

Using the values of *l* and *d* in the first relation, we have *n* = 11.

**Example 7:** The following is known about an AP:

\[{T_m} = \frac{1}{n},\,\,\,{T_n} = \frac{1}{m}\]

Find the value of \({S_{mn}}\).

**Solution:** Let *a* and *d* be the first term and the common difference of the AP. We have:

\[\begin{align}&{T_m} = a + \left( {m - 1} \right)d = \frac{1}{n}\\&{T_n} = a + \left( {n - 1} \right)d = \frac{1}{m}\\& \Rightarrow \,\,\,{T_m} - {T_n} = \left( {m - n} \right)d = \frac{1}{n} - \frac{1}{m}\\& \Rightarrow \,\,\,\left( {m - n} \right)d = \frac{{m - n}}{{mn}}\\ &\Rightarrow \,\,\,d = \frac{1}{{mn}}\end{align}\]

Using this value of *d* in the any of the two initial relations, we obtain \(a = \frac{1}{{mn}}\). Thus,

\[\begin{align}&{S_{mn}} = \left( {\frac{{mn}}{2}} \right)\left( {2a + \left( {mn - 1} \right)d} \right)\\\,\,\,\,\,\,\,\, &\quad\;\;\;= \left( {\frac{{mn}}{2}} \right)\left\{ {\frac{2}{{mn}} + \left( {mn - 1} \right)\left( {\frac{1}{{mn}}} \right)} \right\}\\\,\,\,\,\,\,\,\, &\qquad= \frac{{mn + 1}}{2}\end{align}\]

**Example 8:** The sums of the first *n*, 2*n* and 3*n* terms of an AP are respectively denoted by *x*, *y* and *z*. Find the value of \(\frac{z}{{y - x}}\).

**Solution:** Let *a* and *d* be the first term and the common difference of the AP. We have:

\[\begin{align}&x = {S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\\&y = {S_{2n}} = \frac{{2n}}{2}\left( {2a + \left( {2n - 1} \right)d} \right)\\&z = {S_{3n}} = \frac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)\end{align}\]

Now,

\[\begin{align}&y - x = n\left\{ \begin{array}{l}2a + \left( {2n - 1} \right)d\\ - \frac{1}{2}\left( {2a + \left( {n - 1} \right)d} \right)\end{array} \right\}\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= n\left( {a + \left( {\frac{{3n}}{2} - \frac{1}{2}} \right)d} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;= \frac{1}{3} \times \left\{ {\frac{{3n}}{2}\left( {2a + \left( {3n - 1} \right)d} \right)} \right\} = \frac{z}{3}\\ &\Rightarrow \,\,\,\frac{z}{{y - x}} = 3\end{align}\]

**Example 9:** For an AP,

\[\frac{{{S_m}}}{{{S_n}}} = \frac{{{m^2}}}{{{n^2}}}\]

Find the value of \(\frac{{{T_m}}}{{{T_n}}}\).

**Solution:** We let *a* and *d* be the first term and the common difference of the AP respectively. We have:

\[\begin{align}&\frac{{{S_m}}}{{{S_n}}} = \frac{{{m^2}}}{{{n^2}}}\\ &\Rightarrow \,\,\,\frac{{\frac{m}{2}\left( {2a + \left( {m - 1} \right)d} \right)}}{{\frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)}} = \frac{{{m^2}}}{{{n^2}}}\\ &\Rightarrow \,\,\,\frac{{2a + \left( {m - 1} \right)d}}{{2a + \left( {n - 1} \right)d}} = \frac{m}{n}\end{align}\]

Cross-multiplying and simplifying this will yield \(d = 2a\). Thus,

\[\begin{align}&\frac{{{T_m}}}{{{T_n}}} = \frac{{a + \left( {m - 1} \right)d}}{{a + \left( {n - 1} \right)d}} = \frac{{a + \left( {m - 1} \right)\left( {2a} \right)}}{{a + \left( {n - 1} \right)\left( {2a} \right)}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\qquad\qquad\;\quad= \frac{{2m - 1}}{{2n - a}}\end{align}\]