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Question-31406




Question Number 31406 by rahul 19 last updated on 08/Mar/18
Commented by mrW2 last updated on 08/Mar/18
OM_r =((OA_r +OA_(r+1) )/2)=((1+q)/2)×OA_r   ΣOM_r =((1+q)/2)×ΣOA_r =((1+q)/(2(1−q)))×OA_1
$${OM}_{{r}} =\frac{{OA}_{{r}} +{OA}_{{r}+\mathrm{1}} }{\mathrm{2}}=\frac{\mathrm{1}+{q}}{\mathrm{2}}×{OA}_{{r}} \\ $$$$\Sigma{OM}_{{r}} =\frac{\mathrm{1}+{q}}{\mathrm{2}}×\Sigma{OA}_{{r}} =\frac{\mathrm{1}+{q}}{\mathrm{2}\left(\mathrm{1}−{q}\right)}×{OA}_{\mathrm{1}} \\ $$
Commented by rahul 19 last updated on 09/Mar/18
thank u sir!
$${thank}\:{u}\:{sir}! \\ $$

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