If-log-x-log-y-log-z-x-y-z-gt-1-are-in-GP-then-2x-log-bx-3x-log-by-4x-log-bz-are-in-A-P-True-False-

Question Number 25378 by Tinkutara last updated on 09/Dec/17

I f \log x, \log y, \log z (x, y, z > 1) a r e i n

G P t h e n 2 x + \log (b x), 3 x + \log (b y),

4 x + \log (b z) a r e i n A . P .

T r u e / F a l s e ?

Answered by Rasheed.Sindhi last updated on 09/Dec/17

\frac{\log y}{\log x} = \frac{\log z}{\log y} Given

To be proved :

{3 x + \log (b y)} - {2 x + \log (b x)}

= {4 x + \log (b z)} - {3 x + \log (b y)}

x + \log (b y) - \log (b x) = x + \log (b z) - \log (b y)

x + \log (\frac{b y}{b x}) = x + \log (\frac{b z}{b y})

\log (\frac{y}{x}) = \log (\frac{z}{y})

\log y - \log x = \log z - \log y

∴ \log x, \log y, logz are in AP

We can also prove vice versa

i - e If \log x, \log y, logz are in AP

then 2 x + \dots, 3 x + \dots, 4 x + \dots are

also in AP .

∴ If \log x, \log y, logz are in GP

then 2 x + . ., 3 x + . ., 4 x + \dots {can}^{'} t be

in GP (∵ a sequence {can}^{'} t be both

AP and GP simultneously in general .)

Hence False .

Commented by Tinkutara last updated on 09/Dec/17

Thank you Sir!