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

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(c) $a,\;ar,\;a{r^2}$ are in $G.P.$ $a,\;ar - 8,\;a{r^2} - 64$ are in $A.P.$,

we get $ \Rightarrow $$a({r^2} - 2r + 1) = 64$ .....(i)

Again, $a,

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$4, 20, 100$

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(c) $a,\;ar,\;a{r^2}$ are in $G.P.$ $a,\;ar - 8,\;a{r^2} - 64$ are in $A.P.$,

we get $ \Rightarrow $$a({r^2} - 2r + 1) = 64$ .....(i)

Again, $a,\;ar - 8,\;a{r^2} - 64$ are in $G.P.$

$	herefore $${(ar - 8)^2} = a(a{r^2} - 64)$ or $a(16r - 64) = 64$ .....(ii)

Solving (i) and (ii), we get $r = 5,\;a = 4$. Thus required numbers are $4,20,100.$

Trick : Check by alternates according to conditions

(a) $ \Rightarrow $$4, 20, -28$ which are not in $A.P.$

(b) $ \Rightarrow $$4, 12, -28$ which are also not in $A.P.$

(c) $ \Rightarrow $$4, 20, 36$ which are obviously in $A.P.$ with $16$ as common difference.

These numbers also satisfy the second condition

$i.e.$  $4, 20 -8, 36$ are in $G.P.$

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

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