Accedmychevron_right11thchevron_rightmathchevron_rightSequences And Serieschevron_rightExercise 6.5

(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}) in G.P.

[ \left(\dfrac{1}{b}\right)^2=\dfrac{1}{a}\cdot\dfrac{1}{c}\implies b^2=ac ]

Common ratio (r=\dfrac{1/b}{1/a}=\dfrac{a}{b})

Also (r^2=\dfrac{1/c}{1/a}=\dfrac{a}{c})

[ r=\pm\sqrt{\dfrac{a}{c}} ]

Proved.