Question

The sum of three numbers in G.P. is 70. If the two extremes are multiplied each by 4, and the mean by 5, the products are in A.P. Find the number.

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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.

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.