Answer

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Hint: Take the three numbers as $a,ar,a{{r}^{2}}$ , where $r$ is the common ratio. Then apply the conditions as given.

Complete step-by-step answer:

A G.P. or geometric progression is a sequence of numbers where each new term after the first is obtained by multiplying the preceding term by a constant $r$ called common ratio.

Let us assume that the three numbers are $a,ar,a{{r}^{2}}$, where $r$ is the common ratio.

Now the first condition is the sum of these three numbers is 70. Our first equation will be:

$a+ar+a{{r}^{2}}=70.........(1)$

If we multiply the two extreme terms that is the first term and the third term by 4, we will get:

$\begin{align}

& a\times 4=4a \\

& a{{r}^{2}}\times 4=4a{{r}^{2}} \\

\end{align}$

We know that the geometric mean of these three numbers will be:

$=\sqrt[3]{a\times ar\times a{{r}^{2}}}$

$={{\left( \left( a\times a\times a \right)\times \left( r\times {{r}^{2}} \right) \right)}^{\dfrac{1}{3}}}$

$={{\left( {{a}^{3}}\times {{r}^{3}} \right)}^{\dfrac{1}{3}}}$

$={{\left( {{\left( ar \right)}^{3}} \right)}^{\dfrac{1}{3}}}$

$={{\left( ar \right)}^{3\times \dfrac{1}{3}}}$

$=ar$

If we multiply the geometric mean by 5 we will get:

$ar\times 5=5ar$

Now it is given that these three numbers are in A.P.

$4a,5ar,4a{{r}^{2}}$ are in A.P. So the difference between the 2nd term and the 1st term is the same as the difference between the 3rd term and 2nd term. Now we have:

$5ar-4a=4a{{r}^{2}}-5ar$

$\Rightarrow 5ar+5ar=4a{{r}^{2}}+4a$ , by taking $-5ar$ from right side to left side.

$\Rightarrow 10ar=4a{{r}^{2}}+4a.......(2)$

Multiply both sides of the equation (1) by 4.

$\begin{align}

& 4(a+ar+a{{r}^{2}})=4\times 70 \\

& \Rightarrow 4a+4ar+4a{{r}^{2}}=280 \\

\end{align}$

Now, substitute the value of $4a+4a{{r}^{2}}$ from equation (2).

$\begin{align}

& \Rightarrow 10ar+4ar=280 \\

& \Rightarrow 14ar=280 \\

& \Rightarrow ar=\dfrac{280}{14} \\

& \Rightarrow ar=20 \\

\end{align}$

So, the second number is 20.

Now, substitute this value in equation (1).

$\begin{align}

& a+20+\left( ar \right)\times r=70 \\

& \Rightarrow a+20r=70-20 \\

& \Rightarrow a+20r=50 \\

\end{align}$

Multiply both sides of the equation by $r$ .

$\Rightarrow ar+20{{r}^{2}}=50r$

Substitute the value of $ar$ .

$\Rightarrow 20+20{{r}^{2}}=50r$

$\Rightarrow 20{{r}^{2}}-50r+20=0$ , by taking $50r$ from right side to left side.

Now this is a quadratic equation. We will solve this by middle term factorization.

$\Rightarrow 20{{r}^{2}}-40r-10r+20=0$

$\Rightarrow 20r(r-2)-10(r-2)=0$

Take $\left( r-2 \right)$ common from both the terms.

$\Rightarrow (r-2)(20r-10)=0$

Either,

$\begin{align}

& r-2=0 \\

& \Rightarrow r=2 \\

\end{align}$

Or,

$\begin{align}

& 20r-10=0 \\

& \Rightarrow 20r=10 \\

& \Rightarrow r=\dfrac{10}{20} \\

& \Rightarrow r=\dfrac{1}{2} \\

\end{align}$

If $r=2$ ,

$a=\dfrac{ar}{r}=\dfrac{20}{2}=10$

$a{{r}^{2}}=ar\times r=20\times 2=40$

Therefore the numbers are 10, 20 and 40.

If $r=\dfrac{1}{2}$ ,

$a=\dfrac{ar}{r}=\dfrac{20}{\dfrac{1}{2}}=20\times \dfrac{2}{1}=40$

$a{{r}^{2}}=\left( ar \right)\times r=20\times \dfrac{1}{2}=10$

Therefore the numbers are 40, 20 and 10.

Hence our three numbers are 10, 20 and 40.

Note: Whenever in the question it is given that the numbers are in G.P. always take the numbers as $a,ar,a{{r}^{2}},a{{r}^{3}},...$

Do not take the numbers as $a,b,c,d,....$

It will make the equations more complicated.

Complete step-by-step answer:

A G.P. or geometric progression is a sequence of numbers where each new term after the first is obtained by multiplying the preceding term by a constant $r$ called common ratio.

Let us assume that the three numbers are $a,ar,a{{r}^{2}}$, where $r$ is the common ratio.

Now the first condition is the sum of these three numbers is 70. Our first equation will be:

$a+ar+a{{r}^{2}}=70.........(1)$

If we multiply the two extreme terms that is the first term and the third term by 4, we will get:

$\begin{align}

& a\times 4=4a \\

& a{{r}^{2}}\times 4=4a{{r}^{2}} \\

\end{align}$

We know that the geometric mean of these three numbers will be:

$=\sqrt[3]{a\times ar\times a{{r}^{2}}}$

$={{\left( \left( a\times a\times a \right)\times \left( r\times {{r}^{2}} \right) \right)}^{\dfrac{1}{3}}}$

$={{\left( {{a}^{3}}\times {{r}^{3}} \right)}^{\dfrac{1}{3}}}$

$={{\left( {{\left( ar \right)}^{3}} \right)}^{\dfrac{1}{3}}}$

$={{\left( ar \right)}^{3\times \dfrac{1}{3}}}$

$=ar$

If we multiply the geometric mean by 5 we will get:

$ar\times 5=5ar$

Now it is given that these three numbers are in A.P.

$4a,5ar,4a{{r}^{2}}$ are in A.P. So the difference between the 2nd term and the 1st term is the same as the difference between the 3rd term and 2nd term. Now we have:

$5ar-4a=4a{{r}^{2}}-5ar$

$\Rightarrow 5ar+5ar=4a{{r}^{2}}+4a$ , by taking $-5ar$ from right side to left side.

$\Rightarrow 10ar=4a{{r}^{2}}+4a.......(2)$

Multiply both sides of the equation (1) by 4.

$\begin{align}

& 4(a+ar+a{{r}^{2}})=4\times 70 \\

& \Rightarrow 4a+4ar+4a{{r}^{2}}=280 \\

\end{align}$

Now, substitute the value of $4a+4a{{r}^{2}}$ from equation (2).

$\begin{align}

& \Rightarrow 10ar+4ar=280 \\

& \Rightarrow 14ar=280 \\

& \Rightarrow ar=\dfrac{280}{14} \\

& \Rightarrow ar=20 \\

\end{align}$

So, the second number is 20.

Now, substitute this value in equation (1).

$\begin{align}

& a+20+\left( ar \right)\times r=70 \\

& \Rightarrow a+20r=70-20 \\

& \Rightarrow a+20r=50 \\

\end{align}$

Multiply both sides of the equation by $r$ .

$\Rightarrow ar+20{{r}^{2}}=50r$

Substitute the value of $ar$ .

$\Rightarrow 20+20{{r}^{2}}=50r$

$\Rightarrow 20{{r}^{2}}-50r+20=0$ , by taking $50r$ from right side to left side.

Now this is a quadratic equation. We will solve this by middle term factorization.

$\Rightarrow 20{{r}^{2}}-40r-10r+20=0$

$\Rightarrow 20r(r-2)-10(r-2)=0$

Take $\left( r-2 \right)$ common from both the terms.

$\Rightarrow (r-2)(20r-10)=0$

Either,

$\begin{align}

& r-2=0 \\

& \Rightarrow r=2 \\

\end{align}$

Or,

$\begin{align}

& 20r-10=0 \\

& \Rightarrow 20r=10 \\

& \Rightarrow r=\dfrac{10}{20} \\

& \Rightarrow r=\dfrac{1}{2} \\

\end{align}$

If $r=2$ ,

$a=\dfrac{ar}{r}=\dfrac{20}{2}=10$

$a{{r}^{2}}=ar\times r=20\times 2=40$

Therefore the numbers are 10, 20 and 40.

If $r=\dfrac{1}{2}$ ,

$a=\dfrac{ar}{r}=\dfrac{20}{\dfrac{1}{2}}=20\times \dfrac{2}{1}=40$

$a{{r}^{2}}=\left( ar \right)\times r=20\times \dfrac{1}{2}=10$

Therefore the numbers are 40, 20 and 10.

Hence our three numbers are 10, 20 and 40.

Note: Whenever in the question it is given that the numbers are in G.P. always take the numbers as $a,ar,a{{r}^{2}},a{{r}^{3}},...$

Do not take the numbers as $a,b,c,d,....$

It will make the equations more complicated.

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