Answer:

Using the formula,

a_n = a + (n – 1)d

LHS: a_m+n + a_m-n

a_m+n + a_m-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)] {∵ a_n = a + (n – 1)d}

a_m+n + a_m-n = 2a_m

Thus, Proved.