1. **Problem Statement:** Given an arithmetic progression (A.P.) with terms $a_1, a_2, a_3, \ldots$, and two distinct indices $m \neq n$, it is given that $m$ times the $m$th term equals $n$ times the $n$th term. We need to show that the $(m+n)$th term of the A.P. is zero. 2. **Formula for the $k$th term of an A.P.:** $$a_k = a + (k-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Given condition:** $$m a_m = n a_n$$ Substitute $a_m = a + (m-1)d$ and $a_n = a + (n-1)d$: $$m [a + (m-1)d] = n [a + (n-1)d]$$ 4. **Expand both sides:** $$m a + m (m-1)d = n a + n (n-1)d$$ 5. **Rearrange terms to isolate $a$ and $d$:** $$m a - n a = n (n-1)d - m (m-1)d$$ $$a (m - n) = d [n (n-1) - m (m-1)]$$ 6. **Simplify the right side:** $$n (n-1) - m (m-1) = (n^2 - n) - (m^2 - m) = n^2 - n - m^2 + m = (n^2 - m^2) - (n - m)$$ 7. **Factor the difference of squares:** $$n^2 - m^2 = (n - m)(n + m)$$ So, $$a (m - n) = d [(n - m)(n + m) - (n - m)] = d (n - m) [(n + m) - 1]$$ 8. **Note that $m - n = -(n - m)$, so:** $$a (m - n) = d (n - m) [(n + m) - 1] \implies a (m - n) = -d (m - n) [(n + m) - 1]$$ 9. **Divide both sides by $(m - n)$ (nonzero since $m \neq n$):** $$a = -d [(n + m) - 1] = -d (m + n - 1)$$ 10. **Find the $(m+n)$th term:** $$a_{m+n} = a + ((m+n) - 1)d = a + (m + n - 1)d$$ Substitute $a$ from step 9: $$a_{m+n} = -d (m + n - 1) + (m + n - 1)d = 0$$ **Final answer:** The $(m+n)$th term of the A.P. is zero.