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

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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.

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