sum of the terms of an arithmetic sequence

Relevant wiki: Arithmetic Progressions

The $n^{th}$ term of an arithmetic progression is given by $a_n=a_m+(n-m)d$ where $d$ is the common difference. Using this formula to solve for $d$ , we have

$389=181+(30-14)d$ $\color{#D61F06}\implies$ $d=13$

Using this formula again to solve for $a_1$ , we have

$a_1=a_{14}+(1-14)d=181-13(13)=12$

The sum of the terms of an arithmetic progression is given by $s=\dfrac{n}{2}[2a_1+(n-1)d]$ , we have

$s_{25}=\dfrac{25}{2}[2(12)+24(13)]=\boxed{4200}$

The common difference is 13, so the 13th term is 181 - 13 = 168. The sum of the first 25 terms is 25 times the 13th term, and 25 * 168 = 4200