Three numbers form a GP. If the $$3rd$$ term is decreased by $$64$$, then the three numbers thus obtained will constitute an AP. If the second termof this AP is decreased by $$8$$, a GP will be formed again, then the numbers will be

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Infinity Learn · Accepted Answer

let a, ar, $$ar2$$  are in GP  and a, ar, $$ar2$$ $$-$$ $$64$$ are  in AP, we $$getar2$$−$$2r+1=64$$      …….(i)Again$$, a, ar $$-8$$, $$ar2$$ $$-$$ $$64$$ are  in GP.∴ $$(ar$$−$$8)2=aar2$$−$$64$$⇒ $$a(16r$$−$$64)=64$$     …..$$(ii)On$$ solving  Eqs.$$(i) and (ii), we get r $$=5$$,a $$=4Thus$$,  required numbers  are $$4$$, $$20$$, $$100$$.

Three numbers form a GP. If the 3rd term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second term
of this AP is decreased by 8, a GP will be formed again, then the numbers will be

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Three numbers form a GP. If the 3rd term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second termof this AP is decreased by 8, a GP will be formed again, then the numbers will be

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Three numbers form a GP. If the 3rd term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second term
of this AP is decreased by 8, a GP will be formed again, then the numbers will be