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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