In an A.P., show that am+n + am–n = 2am.

Let common difference of an A.P is d and first term is a


We know, an = a + (n – 1)d where a is first term or a1 and d is common difference


Now, take L.H.S.:


am+n + am-n = a + (m + n – 1)d + a + (m - n – 1)d


am+n + am-n = a + md + nd – d + a + md - nd – d


am+n + am-n = 2a + 2md – 2d


am+n + am-n = 2(a + md – d)


am+n + am-n = 2[a + d(m – 1)]


{ an = a + (n – 1)d}


am+n + am-n = 2am


Hence Proved


2