Find the sum : $18 + 15\dfrac{1}{2} + 13 + …. + \Big(-49\dfrac{1}{2}\Big)$.

The above series is an A.P. with a = 18, d = $13 - 15\dfrac{1}{2} = 13 - \dfrac{31}{2} = \dfrac{26 - 31}{2} = -\dfrac{5}{2}.$

Let nth term be last term so, a_n = $-49\dfrac{1}{2}.$

We know that

a_n = a + (n - 1)d

$\therefore -49\dfrac{1}{2} = 18 + (n - 1) \times -\dfrac{5}{2} \\[1em] \Rightarrow -\dfrac{99}{2} = 18 - \dfrac{5n}{2} + \dfrac{5}{2} \\[1em] \Rightarrow -\dfrac{99}{2} = \dfrac{36 + 5}{2} - \dfrac{5n}{2} \\[1em] \Rightarrow \dfrac{5n}{2} = \dfrac{41}{2} + \dfrac{99}{2} \\[1em] \Rightarrow 5n = 41 + 99 \\[1em] \Rightarrow 5n = 140 \\[1em] \Rightarrow n = 28. \\[1em]$

Using formula S_n = $\dfrac{n}{2}[a + l]$

$\therefore S_{28} = \dfrac{28}{2}\Big[18 + \Big(-\dfrac{99}{2}\Big)\Big] \\[1em] = 14\Big[18 - \dfrac{99}{2}\Big] \\[1em] = 14\Big[\dfrac{36 - 99}{2}\Big] \\[1em] = 14 \times -\dfrac{63}{2} \\[1em] = 7 \times -63 \\[1em] = -441.$

Hence, the sum of the A.P. is -441.