If the sum of the first nn terms of an arithmetic progression is \(\dfrac{3n^2+13n}{2}\), find the 25th term of this progression.
+26367 If the sum of the first n terms of an arithmetic progression is \(\dfrac{3n^2+13n}{2}\),
find the 25th term of this progression.
\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{\dfrac{3n^2+13n}{2}} \\\\ s_{25} &=& s_{24} + a_{25} \\\\ a_{25} &=& s_{25}- s_{24} \\\\ a_{25} &=& \dfrac{3*25^2+13*25}{2}- \dfrac{3*24^2+13*24}{2} \\\\ a_{25} &=& 1100 - 1020 \\ \mathbf{a_{25}} &=& \mathbf{80} \\ \hline \end{array}\)
The 25th term of this progression is \(\mathbf{80}\)
