If 10 times the 10th term of an A.P. is equal to 15 times the 15th term,

Given:

$10 a_{10}=15 a_{15}$

$\Rightarrow 10[a+(10-1) d]=15[a+(15-1) d]$

$\Rightarrow 10(a+9 d)=15(a+14 d)$

$\Rightarrow 10 a+90 d=15 a+210 d$

$\Rightarrow 0=5 a+120 d$

$\Rightarrow 0$

$\Rightarrow 0=a+24 d$

$\Rightarrow a=-24 d \ldots(\mathrm{i})$

To show:

$a_{25}=0$

$\Rightarrow$ LHS : $a_{25}=a+(25-1) d$

$=a+24 d$

$=-24 d+24 d \quad(\operatorname{From}(\mathrm{i}))$

$=0=\mathrm{RHS}$

Hence proved.