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

1.	In an A.P., show that am+n + am–n = 2am.
Answer» Let common difference of an A.P is d and first term is a We know, a_n = a + (n – 1)d Where a is first term or a1 and d is common difference. Now, Take L.H.S.: a_m+n + a_m-n = a + (m + n – 1)d + a + (m - n – 1)d ⇒ a_m+n + a_m-n = a + md + nd – d + a + md - nd – d ⇒ a_m+n + a_m-n = 2a + 2md – 2d ⇒ a_m+n + a_m-n = 2(a + md – d) ⇒ a_m+n + a_m-n = 2[a + d(m – 1)] {∵ an = a + (n – 1)d} ⇒ a_m+n + a_m-n = 2a_m Hence Proved.

Answer»

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

We know,

a_n = a + (n – 1)d

Where a is first term or a1 and d is common difference.

Now,

Take L.H.S.:

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

⇒ a_m+n + a_m-n = a + md + nd – d + a + md - nd – d

⇒ a_m+n + a_m-n = 2a + 2md – 2d

⇒ a_m+n + a_m-n = 2(a + md – d)

⇒ a_m+n + a_m-n = 2[a + d(m – 1)]

{∵ an = a + (n – 1)d}

⇒ a_m+n + a_m-n = 2a_m

Hence Proved.

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