Question

Arithmetic sequences
Question screenshot attached. Please help!

Answer 1

The sum of the first \(40\) terms of the sequence is \(S_{40}=1900\), which means you have
\[\begin{align*}
u_1+u_2+u_3+\cdots+u_{40}&=u_1+(u_1+d)+(u_1+2d)+\cdots+(u_1+39d)\\[1ex]
S_{40}&=40u_1+(1+2+\cdots+39)d\\[1ex]
1900&=40u_1+780d
\end{align*}\]The first equality holds because for any arithmetic sequence \(u_n\), you have \(u_{n+1}=u_n+nd\).

This also means that \(u_{40}=106=u_1+39d\). So all you have to do is solve the system:
\[\begin{cases}
40u_1+780d=1900\\[1ex]u_1+39d=106
\end{cases}\]

Answer 2

Thank you!