Given a m+1 =2a n+1

Prove that : a3m+1=2am+n+1

Let the first term and common difference of the given A.P. be a and d respectively.

Given : am + 1 = 2an + 1

⇒ a + md = 2(a + nd)

Now, a3m + 1 = a + 3md = a + md + 2md = 2(a + nd) + 2nd = 2( a + md + nd) = 2(a + (m + n)d) = 2am + n + 1

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