Question

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.

Answer 1

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.

& \therefore a+ar+a{{r}^{2}}=70 \\\ & a(1+r+{{r}^{2}})=70.......(1) \\\ \end{aligned}$$ It is also given that the extremes are multiplied by 4 and the mean by 5. Here the extremes are $$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{aligned} & 2b=a+c \\\ & a(5ar)=4a+4a{{r}^{2}} \\\ & 10ar=4a+4a{{r}^{2}} \\\ \end{aligned}$$ Let us simplify the above expression by dividing throughout by 2a. $$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{aligned} & 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{aligned}$$ When $$r=2$$, $$a=10.$$ Similarly when $$r={}^{1}/{}_{2}$$, $$\begin{aligned} & 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{aligned}$$ When $$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.