We have already discussed that \({n^{th}}\) term of an arithmetic progression is 

\[a + ( n -1)d\]

Suppose the sum up to \(n\) terms is \(S\), then

\(S = a + (a + d) + (a + 2d) + … + a + (n - 1)d\)

\(S = a + (n - 1)d + a + (n - 2)d + ……… +  a\)

Adding the two equations,

\(2S = \left\{ {2a + (n - 1)d + 2a + (n - 1)d + .....n\;terms} \right\}\)

Or  \(2S = n [ 2a + (n-1) d]\)

\(S = \frac{n}{2}\left[ {2a\, + \,(n - 1)\,d} \right]\)  

This formula can also be written as 

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

\(S = \frac{n}{2}\left[ {a + \ell } \right]\), where \(\ell \) is the last term of the series. 

Sum of the series = \(\frac{n}{2}\left[ {first\,\;term\, + \,last\,\;term} \right]\)

Example 5: Find the maximum number of terms in the given series if every term of this series is positive.

\[1000, 993, 886, ……….\]

Solution: Suppose there are n terms in the given series, first term is 1000 and common difference is - 7.

Hence  \({n^{th}}\)  term  \(= 1000 + ( n - 1) (-7)\)

\(= 1007 - 7n > 0\)

or  \(n < \frac{{1007}}{7} < 143.8\)

Hence, there can be 143 terms in the series.

Example 6: Place 11 terms between 7 and 85 such that they are in AP.

Solution: We have to place 11 terms between 7 and 85. So the total number of terms = 13, for the given AP.

\(a = 7\), last term \(= 85, n = 13\)

Now  \(\ell  = a + (n - 1)d\)

\(\Rightarrow  85 = 7 + (13 - 1) d\)

or  \(12 d = 78\), or \(d = 6.5\)

Hence the series obtained after the insertion of the 11 terms is as follows:

\(7, 13.5, 20, 26.5, …….. 85\)

Example 7:  Find the sum of 70 terms of the series whose \({n^{th}}\)  term is \(\frac{n}{7} + 3\)

Solution: First term of the series  

= \(\frac{1}{7} + 3\) = \(\frac{{22}}{7}\)

second term of the series \(= \frac{2}{7} + 3 = \frac{{23}}{7}\) 

Hence, the common difference = \(\frac{1}{7}\)

Sum up to \({70^{th}}\) term =\(\frac{{70}}{2}\left[ {2 \times \frac{{22}}{7} + 69 \times \frac{1}{7}} \right]\)

\(= 5 [113] = 565\)

Example 8: If \({S_1} = {\rm{ }}\left\{ 1 \right\},{\rm{ }}{S_2} = {\rm{ }}\left\{ {2,{\rm{ }}3} \right\},{\rm{ }}{S_3} = {\rm{ }}\left\{ {4,{\rm{ }}5,{\rm{ }}6} \right\},{S_4} = {\rm{ }}\left\{ {7,{\rm{ }}8,{\rm{ }}9,{\rm{ }}10} \right\}\) and so on, find first term of \({S_{21}}\)

Solution:  \({S_1}\) is containing only 1 term, \({S_2}\) is containing next two terms, \({S_3}\) is containing next three terms and so on. Hence, till \({S_{20}}\) total number of terms used will be:

\(1 + 2 + 3 + 4 + …… 20\) terms \(=\frac{{20(21)}}{2}= 210\).

Hence, \({S_{21}}\) will start with 211.

Example 9: How many terms are common in all the three series \(A, B\) and \(C\), where

\(A = 3 + 9 + 15 + 21 + 27 + ..... 600\) terms

\(B = 1 + 9 + 17 + 25 + 33 + .... 500\) terms

\(C = 9 + 19 + 29 + 39 + 49 + .... 400\) terms

Solution: LCM of the common differences (6, 8, 10) is 120. Hence common difference of the common series will be 120 and first common term in the given series is 9 (by observation).

\({600^{th}}\)  term of the series \(A = {\rm{ }}3{\rm{ }} + {\rm{ }}599 \times 6{\rm{ }} = {\rm{ }}3597\)

\({500^{th}}\) term of the series \(B = {\rm{ }}1{\rm{ }} + {\rm{ }}499 \times 8{\rm{ }} = {\rm{ }}3993\)

\({400^{th}}\) term of the series \(C = {\rm{ }}9{\rm{ }} + {\rm{ }}399 \times 10{\rm{ }} = {\rm{ }}3999\)

Now last term of the common series must be less than the minimum of last terms of series \(A, B\) and \(C\).

\(9 + (n - 1) 120 \le \) lowest last term 

\(9 + (n - 1) 120 \le \)3597 \( \Rightarrow \)n \( \le \)30.9 \( \Rightarrow \)n = 30