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)

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

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