If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.

Let a be the first term and d be the common difference of the AP. Then,

$10 \times a_{10}=15 \times a_{15} \quad$ (Given)

$\Rightarrow 10(a+9 d)=15(a+14 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 2(a+9 d)=3(a+14 d)$

$\Rightarrow 2 a+18 d=3 a+42 d$

$\Rightarrow a=-24 d$

$\Rightarrow a+24 d=0$

$\Rightarrow a+(25-1) d=0$

$\Rightarrow a_{25}=0$

Hence, the 25th term of the AP is 0.