Q. 234.2( 23 Votes )

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: 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


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