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