If the product of three terms of G.P. is 512. | vedclass

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Question

(b) Let three terms of a $G.P.$ are $\frac{a}{r},\;a,\;ar$

So $\frac{a}{r}.\;a.\;ar = 512$

$ \Rightarrow $${a^3} = {8^3}$

$ \Rightarrow $$a =

Vedclass · Accepted Answer

$4, 8, 16$

Vedclass · Answer

(b) Let three terms of a $G.P.$ are $\frac{a}{r},\;a,\;ar$

So $\frac{a}{r}.\;a.\;ar = 512$

$ \Rightarrow $${a^3} = {8^3}$

$ \Rightarrow $$a = 8$

From second condition, we get $\frac{a}{r} + 8,\;a + 6,\;ar$ will be in $A.P.$

$ \Rightarrow $ $2(a + 6) = \frac{a}{r} + 8 + ar$

$ \Rightarrow $$28 = 8\left\{ {\frac{1}{r} + 1 + r} \right\}$

$ \Rightarrow $ $\frac{1}{r} + r + 1 = \frac{7}{2}$

$ \Rightarrow $$\frac{1}{r} + r - \frac{5}{2} = 0$

$ \Rightarrow $ ${r^2} - \frac{5}{2}r + 1 = 0$

$ \Rightarrow $$2{r^2} - 5r + 2 = 0$

$ \Rightarrow $ $(2r - 1)(r - 2) = 0$

$ \Rightarrow $$r = \frac{1}{2},\;r = 2$ $(\because \;r > 1)$

$ \Rightarrow $ $r = 2$.

Hence required numbers are $4,\;8,\;16$.

Trick : Check for $(a)$ $2 + 8,\;4 + 6,\;8$ are not in $A.P.$

$(b)$ $4 + 8,\;8 + 6,\;16\;i.e.\;12,\;14,\;16$ are in $A.P.$

If the product of three terms of $G.P.$ is $512$. If $8$ added to first and $6$ added to second term, so that number may be in $A.P.$, then the numbers are

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