If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is : 

$\text{ Given that the first term of arithmetic progression is }a=3$

$\begin{array}{c}{S}_{25}=S_{40}-S_{25}\\ 2\cdot S_{25}=S_{40}\\ 2\cdot \frac{25}{2}\left[2a+24d\right]=\frac{40}{2}\left[2a+39d\right]\\ 25\left[6+24d\right]=20\left(6+39d\right)\\ 150\left(1+4d\right)=60\left(2+13d\right)\end{array}$

Cancel out the common factor 15 from both sides of the above equation 

$\begin{array}{c}10+40d=8+52d\\ 12d=2\\ d=\frac{1}{6}\end{array}$

$\text{ Therefore, the common difference is }d=\frac{1}{6}$