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  • Need explanation for: Let a1,a2,a3,a4,a5 be a G.P. of positive real numbers such that the A.M. of a2 and a4 is 117 and the G.M. of a2 and a4 is 108.Then the A.M. of a1 and a5 is :

Let a1,a2,a3,a4,a5  be a G.P. of positive real numbers such that the A.M. of  a2 and a4 is 117 and the G.M. of  a2 and a4 is 108.

Then the A.M. of a1 and a5 is :

  • Option 1)

    145.5

  • Option 2)

    108

  • Option 3)

    117

  • Option 4)

    144.5

 

Answers (1)

best_answer

As we learnt in 

Common ratio of a GP (r) -

The ratio of two consecutive terms of a GP

- wherein

eg: in 2, 4, 8, 16, - - - - - - -

r = 2

and in 100, 10, 1, 1/10 - - - - - - -

r = 1/10

 

Let G1=G

so that G2=Gr

           G3=Gr2

               G4=Gr3

               G5=Gr4

 

If G1, G2, G3, G4, G5, in G.P.

G2+G4=234

G2G4=(108)2

Then \frac{G1+G5}{2}= ?

 

So that G_{2}+G_{4}= Gr+Gr^{3}= 234

                           = Gr\left ( 1+r^{2} \right )= 234                 (i)

 

and      Gr\times Gr^{3}= \left ( 108 \right )^{2}

           G^{2}r^{4}= \left ( 108 \right )^{2}

\therefore Gr^{2}= 108                        (ii)

       from (i) and (ii)

    \frac{108}{r^{2}}\times r(17r^{2})= 234

    \therefore 54r^{2}-117r+54=0

    \therefore r= \frac{2}{3} \: \: so \: a= 243

     \therefore \frac{a_{1}+a_{5}}{2}

        = \frac{a+ar^{4}}{2}

        =\frac{a}{2}\left ( 1+r^{4} \right )= \frac{27\times 9}{2}\left [ 1+\left ( \frac{2}{3} \right )^{4} \right]

                              =\frac{243}{2}\left ( \frac{81+16}{81} \right )

                              =\frac{243}{2}\times \frac{97}{81}

                              =145.5

 

 


Option 1)

145.5

This option is correct

Option 2)

108

This option is incorrect

Option 3)

117

This option is incorrect

Option 4)

144.5

This option is incorrect

Posted by

Plabita

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