Statement-I     The 25^th term of an A.P $-5,\frac{-5}{2},0,\frac{5}{2},\cdots$  is 70.

$\mathbf{Statement}\mathbf{-}\mathbf{II} \mathrm{In} \mathrm{an} \mathrm{AP}, \mathrm{if} {\mathrm{a}}_{16}=\mathrm{P}, \mathrm{then} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{first} 31 \mathrm{terms} \mathrm{of} \mathrm{it} \mathrm{is} 16\mathrm{Px}.$

Here $a=-5,d=\frac{5}{2},n=25$

$T_n=a+\left(n-1\right)d$

$T_{25}=-5+12\left(5\right)=-5+60=55$

Statement(i) is false

Statement(ii): Given that $a_{16}=p\Rightarrow a+15d=p$

Now $s_{31}=\frac{31}{2}\left[2a+30d\right]$

$=\frac{31}{2}\left[2\left(a+15d\right)\right]$

$=31\left(a+15d\right)$

$=31p$