tn = a + (n – 1)d
Given tm + 1 = 2 tn + 1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md = 2a + 2nd
md – 2nd = a
d(m – 2n) = a … (1)
To Prove t(3m + 1) = 2(tm + n + 1)
L.H.S. = t3m + 1
= a + (3m + 1 – 1)d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)
R.H.S. = 2(tm + n + 1)
= 2 [a + (m + n + 1 – 1)d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2 md – nd]
= 2d (2m – n)
R.H.S = L.H.S
∴ t(3m + 1) = 2 t(m + n + 1)
Hence it is proved.