If in an A.P., T_m = n, T_n = m, prove d = –1.

Formula Used.

a_n = a + (n–1)d

a_m = a + (m–1)d

If m^th term of A.P is given as n

Then,

a + (m–1)d = n

we get a = n–(m–1)d ......eq 1

a_n = a + (n–1)d

If n^th term of A.P is given as m

Then,

a + (n–1)d = m

we get a = m–(n–1)d ......eq 2

Equating both eq 1 and eq 2

We get ;

n–(m–1)d = m–(n–1)d

(n–1)d–(m–1)d = m–n

(n–1–(m–1))d = m–n

(n–1–m + 1)d = m–n

(n–m)d = –(n–m)

d = = –1