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