Question

Find the sum up to $30$ terms of an $AP$ whose second term is $\frac{1}{2}$ and ${29}^{th}$ term is $49\frac{1}{2}$.

Answer 1

$t_2=\frac{1}{2}=>a+\left(1\right)d=\frac{1}{2}...........\left(1\right)$

$t_{29}=49\frac{1}{2}=>a+28d=49\frac{1}{2}..............\left(2\right)$

$=>a=\frac{1}{2}-d$

$=>\frac{1}{2}-d+28d=49\frac{1}{2}$

$=>\frac{1}{2}+27d=\frac{99}{2}$

$=>1+54d=99$

$=>d=\frac{98}{54}=\frac{49}{27}$

$S_{30}=\frac{30}{2}[2\times (\frac{-71}{54})+29\times \frac{49}{27}]$

$=15[\frac{29\times 49\times 2}{27\times 2}-\frac{142}{54}]$

$=15\times \frac{2700}{54}$

$=750$