If $10$ times the ${10^{th}}$term of an A.P. is equal to $15$ times the ${15^{th}}$ term, show that the ${25^{th}}$term of A.P. is zero.

Hint: Use general term of A.P. i.e, ${T_n} = a + (n - 1)d$.
We, know that the ${n^{th}}$term of an A.P. is given as:
${T_n} = a + (n - 1)d$
$\therefore {10^{th}}$term of A.P. will be:
$
   \Rightarrow {T_{10}} = a + (10 - 1)d \\
   \Rightarrow {T_{10}} = a + 9d \\
$
Similarly, ${15^{th}}$term will be:
$
   \Rightarrow {T_{15}} = a + (15 - 1)d, \\
   \Rightarrow {T_{15}} = a + 14d \\
$
Now, according to question:
$10{T_{10}} = 15{T_{15}}$
So, putting values of ${T_{10}}$and ${T_{15}}$from above, we’ll get:
$
   \Rightarrow 10(a + 9d) = 15(a + 14d) \\
   \Rightarrow 10a + 90d = 15a + 210d \\
   \Rightarrow 5a + 120d = 0 \\
   \Rightarrow a + 24d = 0 \\
$
And ${25^{th}}$term of A.P. will be:
$
   \Rightarrow {T_{25}} = a + (25 - 1)d \\
   \Rightarrow {T_{25}} = a + 24d \\
$
Putting the value $a + 24d = 0$ from above, we get:
$ \Rightarrow {T_{25}} = 0.$
Hence the ${25^{th}}$term of A.P. is zero.
Note: Since ${25^{th}}$ term of A.P. is zero, we can conclude that the sum of the first 49 terms of this A.P. is zero. In that case, the sum of the first 24 terms will be negative of the sum of the last 24 terms and ${25^{th}}$ term is already zero.