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

Here, let us take the first term of the A.P. as a and the common difference as d

We are given that 10 times the 10th term is equal to 15 times the 15th term. We need to show that 25th term is zero.

So, let us first find the two terms.

So, as we know,

$a_{n}=a+(n-1) d$

For $10^{\text {th }}$ term $(n=10)$,

$a_{10}=a+(10-1) d$

$=a+9 d$

For $15^{\text {th }}$ term $(n=15)$,

$a_{15}=a+(15-1) d$

$=a+14 d$

Now, we are given,

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

Solving this, we get,

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

$90 d-210 d=15 a-10 a$

$-120 d=5 a$

$-24 d=a$ .......(1)

Next, we need to prove that the 25th term of the A.P. is zero. For that, let us find the 25th term using n = 25,

$a_{25}=a+(25-1) d$

$=-24 d+24 d \quad($ Using 1$)$

$=0$

Thus, the 25th term of the given A.P. is zero.

Hence proved