Three numbers are in G.P. whose sum is 70. If the extremes are each multiplied by 4 and the means by 5, they will be in A.P. Find the numbers.

Hint: Assume 3 numbers of G.P. as \[a,ar,a{{r}^{2}}\]. Find an expression related to it. Then find the 3 numbers that are in A.P. and find an equation using rule \[2b=a+c.\]

Complete step-by-step answer:
Solve the quadratic equation formed and find the numbers by substituting the roots in the series.
A geometric progression is a sequence in which each term is derived by multiplication or dividing the preceding term by a fixed number called the common ratio, which is denoted by ‘r’. Let the first term be denoted as ‘a’.
Let us consider the three numbers as \[a,ar,a{{r}^{2}}.\]
It is given that the sum of three numbers is 70.
\[\begin{align}
  & \therefore a+ar+a{{r}^{2}}=70 \\
 & a(1+r+{{r}^{2}})=70.......(1) \\
\end{align}\]
It is also given that the extremes are multiplied by 4 and the mean by 5. Here the extremes are \[a\] and \[a{{r}^{2}}.\] So they become \[4a\] and \[4a{{r}^{2}}\], and the mean becomes \[5ar.\]
It is said that \[4a,5ar,4a{{r}^{2}}\] forms an A.P.
If 3 terms are in A.P. [ for e.g.- a, b, c are in A.P., then they can be written as \[2b=a+c\]]. Now let us apply this concept in our given A.P.
Here \[a=4a,b=5ar,c=4a{{r}^{2}}.\]
\[\begin{align}
  & 2b=a+c \\
 & a(5ar)=4a+4a{{r}^{2}} \\
 & 10ar=4a+4a{{r}^{2}} \\
\end{align}\]
Let us simplify the above expression by dividing throughout by 2a.
\[5r=2+2{{r}^{2}}\].
Let us rearrange the above expression.
\[2{{r}^{2}}-5r+2=0.\]
The given equation is similar to the general quadratic equation \[a{{x}^{2}}+bx+c=0.\]
Let us compare these equations. We get, \[a=2,b=-5,c=2.\]
Let us apply these values in quadratic formula, \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\]
\[\dfrac{-(-5)\pm \sqrt{{{(-5)}^{2}}-4\times 2\times 2}}{2\times 2}=\dfrac{5\pm \sqrt{25-16}}{4}=\dfrac{5\pm \sqrt{9}}{4}=\dfrac{5\pm 3}{4}.\]
Hence the roots of r are \[\left( \dfrac{5+3}{4} \right)\] and \[\left( \dfrac{5-3}{4} \right)\] \[=2\] and \[{}^{1}/{}_{2}\].
So we got \[r=2\] and \[r={}^{1}/{}_{2}.\]
Let us put \[r=2\] in equation (1) and find \[a\].
\[\begin{align}
  & a(1+r+{{r}^{2}})=70 \\
 & a(1+2+{{2}^{2}})=70 \\
 & a(1+2+4)=70 \\
 & a\times 7=70 \\
 & \Rightarrow a={}^{70}/{}_{7}=10. \\
\end{align}\]
When \[r=2\], \[a=10.\]
Similarly when \[r={}^{1}/{}_{2}\],
\[\begin{align}
  & a\left( 1+\dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}} \right)=70 \\
 & a\left( 1+\dfrac{1}{2}+\dfrac{1}{4} \right)=70 \\
 & a\left[ \dfrac{4+2+1}{4} \right]=70 \\
 & \Rightarrow a=\dfrac{70\times 4}{7}=40. \\
\end{align}\]
When \[r={}^{1}/{}_{2}\], \[a=40.\]
Thus when \[r=2\] and \[a=10\] the G.P. \[a,ar,a{{r}^{2}}\] becomes,
\[10,(10\times 2),\left( 10\times {{2}^{2}} \right)=10,20,40.\]
When \[r={}^{1}/{}_{2}\] and \[a=40\]the G.P. \[a,ar,a{{r}^{2}}\] becomes,
\[40,\left( 40\times {}^{1}/{}_{2} \right),\left( 40\times {}^{1}/{}_{4} \right)=40,20,10.\]
Hence the numbers are 10, 20, 40 or 40, 20, 10.

Note: It is important to know the basic formulae of G.P. and A.P. We can only find the value of r with the help of a second equation formed by 3 terms of A.P. We should be able to understand this concept of A.P. from the question when it is given with terms like extremes and mean.