Question

Will give brainliest

Answer 1

$\displaystyle\sum_{n=2}^{91}(3n+8)=\sum_{n=1}^{91}(3n+8)-a_1=(*)\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\a_1=3\cdot1+8=11\\n=91\\a_n=3n+8\\d=a_n-a_{n-1}\\a_2=3\cdot2+8=14\\d=14-11=3\\a_{91}=11+(91-1)\cdot 3=11+90\cdot3=281\\\\S_{91}=\dfrac{11+281}{2}\cdot91=146\cdot91=13286\\\\(*)=13286-11=\boxed{13275}$