In the four numbers first three are in G.P. a | vedclass

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Question

(d) Let the numbers be $\frac{a}{r},\;a,\;ar,\;2ar - a$&hellip;..$(i)$

Where first three numbers are in $G.P.$ and last three are in $A.P.$

Give

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(d) Let the numbers be $\frac{a}{r},\;a,\;ar,\;2ar - a$&hellip;..$(i)$

Where first three numbers are in $G.P.$ and last three are in $A.P.$

Given that the common difference of $A.P.$ is $6$, so

$ar - a = 6$&hellip;..$(ii)$

Also given $\frac{a}{r} = 2ar - a $

$\Rightarrow \frac{a}{r} = 2\,(ar - a) + a$

$ \Rightarrow $ $\frac{a}{r} = 2(6) + a,$ from $(ii)$

$ \Rightarrow $ $\left( {\frac{a}{r}} \right) - a = 12$

$ \Rightarrow $$a(1 - r) = 12r$

$ \Rightarrow $$r = - \frac{1}{2}$

From $(i)$ we get, $a\left[ {\left( { - \frac{1}{2}} \right) - 1} \right] = 6$ or $a = - 4$

Required numbers from $(i)$ are $8, - 4,\;2,\;8$ .

In the four numbers first three are in $G.P.$ and last three are in $A.P.$ whose common difference is $6$. If the first and last numbers are same, then first will be

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