Three numbers form a G.P. If the 3^{rd} term | vedclass

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(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$Therefore,$2(ar) = a + (ar^2 - 6

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$4, 20, 100$

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(C) Let the three numbers in $G.P.$ be $a, ar, ar^2$.According to the first condition,$a, ar, ar^2 - 64$ are in $A.P.$Therefore,$2(ar) = a + (ar^2 - 64)$ $\Rightarrow a(r^2 - 2r + 1) = 64$ $\Rightarrow a(r - 1)^2 = 64$ .....$(i)$According to the second condition,$a, ar - 8, ar^2 - 64$ are in $G.P.$Therefore,$(ar - 8)^2 = a(ar^2 - 64)$ $\Rightarrow a^2r^2 - 16ar + 64 = a^2r^2 - 64a$ $\Rightarrow 64 - 16ar = -64a$ $\Rightarrow 16ar - 64a = 64$ $\Rightarrow ar - 4a = 4$ .....(ii)From $(i)$,$a(r-1)^2 = 64$. From (ii),$a(r-4) = 4 \Rightarrow a = \frac{4}{r-4}$.Substituting $a$ in $(i)$: $\frac{4}{r-4}(r-1)^2 = 64$ $\Rightarrow (r-1)^2 = 16(r-4)$ $\Rightarrow r^2 - 2r + 1 = 16r - 64$ $\Rightarrow r^2 - 18r + 65 = 0$.Solving for $r$: $(r-5)(r-13) = 0$,so $r=5$ or $r=13$.If $r=5$,$a = \frac{4}{5-4} = 4$. The numbers are $4, 4(5), 4(5^2) = 4, 20, 100$.Checking the conditions: $4, 20, 100-64=36$ are in $A.P.$ $(20-4=16, 36-20=16)$.$4, 20-8=12, 36$ are in $G.P.$ $(12/4=3, 36/12=3)$.Thus,the numbers are $4, 20, 100$.

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

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