In an $A.P.$,ten times the $10^{th}$ term equals fifteen times the $15^{th}$ term. Prove that the $25^{th}$ term of the $A.P.$ is $0$.

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n - 1)d$.
According to the problem,$10 \times a_{10} = 15 \times a_{15}$.
Substituting the formula for the terms: $10(a + 9d) = 15(a + 14d)$.
Divide both sides by $5$: $2(a + 9d) = 3(a + 14d)$.
Expand the brackets: $2a + 18d = 3a + 42d$.
Rearrange the terms: $2a - 3a = 42d - 18d$.
$-a = 24d$,which implies $a = -24d$.
Now,we need to find the $25^{th}$ term,$a_{25} = a + (25 - 1)d = a + 24d$.
Substitute $a = -24d$ into the expression: $a_{25} = -24d + 24d = 0$.
Thus,the $25^{th}$ term of the $A.P.$ is $0$.