If the last term of an A.P. of 30 terms is 119 and the 8^th term from the end (towards the first term) is 91, then find the common difference of the…

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 \[a=3 \text { and } d=4 \] \(\displaystyle \therefore S_{30}=\frac{n}{2}(a+l)\) \[=\frac{30}{2}(3+119) \] = $\displaystyle 1830$