While assuming the terms we should keep one thing in our mind that assumption of the terms should be such which makes the calculations easy.

Whenever we have to assume odd number of terms in \(AP\), then the common difference is taken as \(‘d’\) and whenever we have to assume even number of terms, then the common difference is taken as \(2d\).

Three terms in \(AP = a - d, a, a + d\)

Four terms in \(AP = a - 3d, a - d, a + d, a +3d\)

Five terms in \(AP = a - 2d, a  - d, a, a + d, a + 2d\).

Example: The interior angles of a pentagon are in an arithmetic progression (AP), and the largest angle measures 116°. Find all the angles of the pentagon.

Solution: Let the angles of the pentagons are \(a - 2d,\;a - d,\,\,a,\,\,a + d,\,\,a + 2d\), then sum of the internal angles = \(5a\).

As we know the sum of the internal angles of a polygon  is \((n - 2) \times 180\), so the sum of the internal angles of a pentagon = \((5 - 2) \times 180 = 540\)

Hence \(5a = 540 \Rightarrow a = 108\)

The biggest angle \(=a+2d= 116\), hence \(d=4\)

The angles are \(100,\;104,\;108,\;112,\;116\)

Example 10: The first and the last two terms of an \(A.P.\) are \(a, b, c\) respectively. Find its sum.

Solution: The common difference of the given \(A.P.\) is 

\(d = c – b\)

Let the number of terms in the \(A.P.\) be \(n\), then we have

\(c = a + (n - 1)d = a + (n - 1)(c - b)\)

Gives \(n = \frac{{c - a}}{{c - b}} + 1 = \frac{{2c - a - b}}{{c - b}}\)

Hence, the required sum is given by 

\(S = \frac{n}{2}(a + c) = \frac{{(2c - a - b)(a + c)}}{{2(c - b)}}\)

Example 11: If the \({m^{th}}\) term of an \(A.P.\) is \(\frac{1}{n}\) and the \({n^{th}}\) term is \(\frac{1}{m}\), then find the \({\left( {mn} \right)^{th}}\) term.

Solution: Let \(a\) and \(d\) be the first term and the common difference of the \(A.P.\) respectively.

Then we have

\(a + (m - 1)d = \frac{1}{n}\) .…(1)

and \(a + (n - 1)d = \frac{1}{m}\) .…(2)

Subtracting equation (2) from equation (1), we have

\( (m - n)d = \frac{1}{n} - \frac{1}{m}\) gives \(d = \frac{1}{{mn}}\)

Putting in equation (1), we have \(a = \frac{1}{{mn}}\)

Thus, the \({\left( {mn} \right)^{th}}\) term is 

\(= a + (mn - 1)d = \frac{1}{{mn}} + (mn - 1) \) \(\frac{1}{{mn}} = 1\)

Example 12: If sum up to \(3n\) terms of an AP is given by \({S_{3n}} = 84\), find the value of \({S_{2n}} - {S_n}\)

Solution: Suppose first term and common difference of the series are \(a\) and \(d\) respectively.

\({S_{3N}}\) = \(\frac{{3N}}{2}\left[ {2a + (3N - 1)d} \right] = 84\)………. (1) 

Also \({S_{2N}}\) = \(\frac{{2N}}{2}\left[ {2a + \left( {2N - 1} \right)d} \right]\)……… (2)

\({S_N}\) =\(\frac{N}{2}\left[ {2a + \left( {N - 1} \right)d} \right]\)         …………. (3)

From (2) and (3) 

\({S_{2N}}--{S_N} = \) \(\frac{N}{2}\left[ {2a + (3N - 1)d} \right]\)  \( = \frac{1}{3} \times {S_{3N}} = 28\)