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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Updated on: 10-Oct-2022

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