# If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Given: am+1 = 2an+1

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

Now,

an = a + (n – 1)d

am+1 = a + (m + 1 – 1)d

am+1 = a + md

and an+1 = a + (n + 1 – 1)d

an+1 = a + nd

Given: am+1 = 2an+1

a +md = 2(a + nd)

a + md = 2a + 2nd

md – 2nd = 2a – a

d(m – 2n) = a …(i)

Now,

am+n+1 = a + (m + n + 1 – 1)d

= a + (m + n)d

= md – 2nd + md + nd [from (i)]

= 2md – nd

am+n+1 = d (2m – n) …(ii)

a3m+1 = a + (3m + 1 – 1)d

= a + 3md

= md – 2nd + 3md [from (i)]

= 4md – 2nd

= 2d( 2m – n)

a3m+1 = 2am+n+1 [from (ii)]

Hence Proved

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