# If (m + 1)th

Given: (m + 1)th term of an A.P. is twice the (n + 1)th term

am+1 = 2an+1

To prove: a3m+1 = 2am+n+1

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

When n = m + 1:

am+1 = a + (m + 1 – 1)d

am+1 = a + md

When n = n + 1:

an+1 = a + (n + 1 – 1)d

an+1 = a + nd

According to question:

am+1 = 2an+1

a + md = 2(a + nd)

a + md = 2a + 2nd

a – 2a + md – 2nd = 0

-a + (m – 2n)d = 0

a = (m – 2n)d………(i)

an = a + (n – 1)d

When n = m + n + 1:

am+n+1 = a + (m + n + 1 – 1)d

am+n+1 = a + md + nd

am+n+1 = (m – 2n)d + md + nd……… (From (i))

am+n+1 = md – 2nd + md + nd

am+n+1 = 2md – nd………(ii)

When n = 3m + 1:

a3m+1 = a + (3m + 1 – 1)d

a3m+1 = a + 3md

a3m+1 = (m – 2n)d + 3md……… (From (i))

a3m+1 = md – 2nd + 3md

a3m+1 = 4md – 2nd

a3m+1 = 2(2md – nd)

a3m+1 = 2am+n+1…………(From (ii))

Hence Proved

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