If first term of an A.P. is ‘\(p\)’, second term is ‘\(q\)’ and last term is ‘\(r\)’, then show that the sum of all terms is \(\dfrac{(p + r)(q + r - 2p)}{2(q - p)}\).

Answer (proof): first term \(a = p\), second term \(t_2 = q\), so \(d = q - p\).

Let the A.P. have \(n\) terms. The last term is \(r\):

\(r = a + (n - 1)d = p + (n - 1)(q - p)\)

\(\therefore\ (n - 1)(q - p) = r - p \Rightarrow n - 1 = \dfrac{r - p}{q - p} \Rightarrow n = \dfrac{r - p + q - p}{q - p} = \dfrac{q + r - 2p}{q - p}\).

Sum of all terms \(S_n = \dfrac{n}{2}(a + r) = \dfrac{n}{2}(p + r)\)

\(= \dfrac{1}{2} \cdot \dfrac{q + r - 2p}{q - p} \cdot (p + r) = \dfrac{(p + r)(q + r - 2p)}{2(q - p)}\), as required.