**Solution:**

**mt**

_{m}= nt_{n }(Given)**∴**

**m [a + (m – 1) d] = n [a + (n – 1) d]**

**[**

**∵**

**t**

_{n}= a + (n – 1) d]

**∴**

**m [a + md – d] = n [a + nd – d]**

**∴**

**am + m**

^{2}d – md = an + n^{2}d – nd

**∴**

**am – an + m**

^{2}d – n^{2}d – md + nd = 0

**∴**

**a (m – n) + d (m**

^{2}– n^{2}) – d (m – n) = 0

**∴**

**a (m – n) + d (m + n) (m – n) – d (m – n) = 0**

**[**

**∵**

**a**

^{2}– b^{2}= (a + b) (a – b)]

**Dividing through at by m – n, we get**

**a + d (m + n) - d = 0**

**∴**

**a + d [m + n – 1] = 0**

**∴**

**a + (m + n – 1) d = 0 ------------------------eq. no. (1)**

**t**

_{m+n}= a + (m + n – 1) [**∵**

**t**

_{n}= a + (n – 1) d]

**∴**

**t**

_{m + n}= 0 [from eq. no. (1)]

**Hence proved.**