Question:
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Solution:
Given:
$a_{24}=2 a_{10}$
$\Rightarrow a+(24-1) d=2[a+(10-1) d]$
$\Rightarrow a+23 d=2(a+9 d)$
$\Rightarrow a+23 d=2 a+18 d$
$\Rightarrow 5 d=a \quad \ldots$ (i)
To prove:
$a_{72}=2 a_{34}$
LHS : $a_{72}=a+(72-1) d$
$\Rightarrow a_{72}=a+71 d$
$\Rightarrow a_{72}=5 d+71 d \quad(\operatorname{From}(\mathrm{i}))$
$\Rightarrow a_{72}=76 d$
RHS : $2 a_{34}=2[a+(34-1) d]$
$\Rightarrow 2 a_{34}=2(a+33 d)$ (From (i))
$\Rightarrow 2 a_{34}=2(38 d)$
$\Rightarrow 2 a_{34}=76 d$
$\therefore \mathrm{RHS}=\mathrm{LHS}$
Hence, proved.
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