1.

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.

Answer»

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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