If (m + 1)th term of an A.P. is twice the (n + 1)th term,

Question:

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.

Solution:

Given:

$a_{m+1}=2 a_{n+1}$

$\Rightarrow a+(m+1-1) d=2[a+(n+1-1) d]$

$\Rightarrow a+m d=2(a+n d)$

$\Rightarrow a+m d=2 a+2 n d$

$\Rightarrow 0=a+2 n d-m d$

$\Rightarrow n d=\frac{m d-a}{2} \quad \ldots(\mathrm{i})$

To prove:

$a_{3 m+1}=2 a_{m+n+1}$

LHS : $a_{3 m+1}=a+(3 m+1-1) d$

$\Rightarrow a_{3 m+1}=a+3 m d$

RHS : $2 a_{m+n+1}=2[a+(m+n+1-1) d]$

$\Rightarrow 2 a_{m+n+1}=2(a+m d+n d)$

$\Rightarrow 2 a_{m+n+1}=2\left[a+m d+\left(\frac{m d-a}{2}\right)\right] \quad($ From $(\mathrm{i}))$

$\Rightarrow 2 a_{\mathrm{m}+\mathrm{n}+1}=2\left[\frac{2 a+2 m d+m d-a}{2}\right]$

$\Rightarrow 2 a_{m+n+1}=2\left[\frac{a+3 m d}{2}\right]$

$\Rightarrow 2 a_{m+n+1}=a+3 m d$

∴ LHS = RHS

 

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