If the ratio of sums to n terms of two A.P.’s is $(5n+7) : (3n+2),$ then the ratio of their $17th$ terms is

Let two A.P.'s be
      $a,a+d,a+2d,a+3d,\dots$

and $A,A+D,A+2D,A+3D,\dots$
According to the given condition, 
$\frac{\frac{n}{2}[2a+(n-1\left)d\right]}{\frac{n}{2}[2A+(n-1\left)D\right]}=\frac{5n+7}{3n+2} \Rightarrow \frac{a+\frac{n-1}{2}d}{A+\frac{n-1}{2}D}=\frac{5n+7}{3n+2}$
Putting $\frac{n-1}{2}=16$ or $n=33,$we get 
      $\frac{a+16d}{A+16D}=\frac{5\left(33\right)+7}{3\left(33\right)+2}=\frac{172}{101}$