If the last term of an A.P. of $\displaystyle 30$ terms is $\displaystyle 119$ and the $\displaystyle 8^{\text {th }}$ term from the end (towards the first term) is $\displaystyle 91$, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.

Last term \(\displaystyle a_{n}=119 \Rightarrow \mathrm{a}+29 \mathrm{~d}=119\) \(\displaystyle 8^{\text {th }}\) term from end \(\displaystyle =23^{\text {rd }}\) term from the beginning \[\Rightarrow a+22 d=91 \] Solving (i) and (ii), we get \[a=3 \text { and } d=4 \] \(\displaystyle \therefore S_{30}=\frac{n}{2}(a+l)\) \[=\frac{30}{2}(3+119) \] = $\displaystyle 1830$