find the common difference of an AP

The $n$ th term of an arithmetic progression is given by $a_n = a_1 + (n-1)d$ , where $a_1$ is the first term and $d$ , the common difference. Therefore,

$\begin{cases} a_1 + 49d = 137.5 & ...(1) \\ a_1 + 33d = 97.5 &...(2) \end{cases}$ $\implies (1) - (2): 16 d = 40 \implies d = \dfrac {40}{16} = \boxed{2.5}$

Given in the problem: $a_{50}=137.5$ and $a_{34}=97.5$

Working formula: $a_n=a_m+(n-m)(d)$

Solution:

$a_{50}=a_{34}+(50-34)(d)$

$137.5=97.5+16d$

$d=\boxed{2.5}$