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  • State whether statement in True or False. The sum or difference of two G.P.s, is again a G.P.

State whether statement in True or False.  The sum or difference of two G.P.s, is again a G.P.

Answers (1)

Let the two G.P be a,ar_{1},ar_{1}^{2},ar_{1}^{3}, \ldots

And b,br_{2},br_{2}^{2},br_{2}^{3}, \ldots \\\\

The terms of sum of these two G.P are     

    \\T_{1}= \left( b+a \right) \\\\ T_{2}= \left( br_{2}+ar_{1} \right) \\\\ T_{3}= \left( br_{2}^{2}+ar_{1}^{2} \right) \\\\ r^{'}=\frac{T_{3}}{T_{2}}=\frac{T_{2}}{T_{1}} \\\\
   \\ \frac{T_{3}}{T_{2}}=\frac{br_{2}^{2}+ar_{1}^{2}}{br_{2}+ar_{1}}~and~\frac{T_{2}}{T_{1}}=\frac{br_{2}+ar_{1}}{b+a} \\\\ \frac{T_{3}}{T_{2}} \neq \frac{T_{2}}{T_{1}} \\\\

 

The terms of difference of two G.P are     


  \\ T_{1}= \left( b-a \right) \\\\ T_{2}= \left( br_{2}-ar_{1} \right) \\\\ T_{3}= \left( br_{2}^{2}-ar_{1}^{2} \right) \\\\ r^{'}=\frac{T_{3}}{T_{2}}=\frac{T_{2}}{T_{1}} \\\\
    \\ \frac{br_{2}^{2}-ar_{1}^{2}}{br_{2}-ar_{1}}=\frac{br_{2}-ar_{1}}{b-a} \\\\ \frac{T_{3}}{T_{2}} \neq \frac{T_{2}}{T_{1}} \\\\

Hence, the difference and sum of 2 G.P is not a G.P because common ratio is not same.   

 

 

 

Posted by

infoexpert21

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