Answer:
Using the formula,
an = a + (n – 1)d
LHS: am+n + am-n
am+n + am-n = a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a + md – nd – d
= 2a + 2md – 2d
= 2(a + md – d)
= 2[a + d(m – 1)] {∵ an = a + (n – 1)d}
am+n + am-n = 2am
Thus, Proved.