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.
Given:
$(m + 1)$th term of an A.P. is twice the $(n + 1)$th term.
To do:
We have to prove that $(3m + 1)$th term is twice the $(m + n + 1)$th term.
Solution:
Let the required A.P. be $a, a+d, a+2d, ......$
Here,
$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$
We know that,
$a_n=a+(n-1)d$
Therefore,
$a_{(m+1)}=a+(m+1-1)d$
$=a+md$
$a_{(n+1)}=a+(n+1-1)d$
$=a+nd$
According to the question,
$a_{(m+1)}=2\times a_{(n+1)}$
$a+md=2(a+nd)$
$a+md=2a+2nd$
$2a-a=md-2nd$
$a=(m-2n)d$......(i)
$a_{(3m+1)}=a+(3m+1-1)d$
$=a+3md$
$=md-2nd+3md$ (From (i))
$=4md-2nd$
$=2d(2m-n)$....(ii)
$a_{(m+n+1)}=a+(m+n+1-1)d$
$=a+(m+n)d$
$=md-2nd+md+nd$
$=2md-nd$
$=d(2m-n)$.....(iii)
From (ii) and (iii), we get,
$a_{(3m+1)}=2\times a_{(m+n+1)}$
Hence proved.
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