Question

The sum of two numbers is 6 times their geometric mean, then show that numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$ .

Answer 1

Hint: To solve this question, first we will write the data given in equation form by letting two numbers be a and b. then we will evaluate the values of a + b and a – b. After that using componendo and dividendo, we will find the ratios of two numbers, which are a and b.

Complete step-by-step answer:
Before we solve this question let us see what is an A.P and G.P.
A.P. is a special series of form a, a + d, a + 2d,….., a + ( n -1 )d where a is first term of an A.P and d is common difference between two consecutive terms.
${{n}^{th}}$ term of an A.P. is determined by ${{a}_{n}}=a+(n-1)d$ and summation of n terms of an A.P is $\dfrac{n}{2}(2a+(n-1)d)$ and also if a, b, c are in A.P and d is common difference of an A.P then,
 b - a = c – b = d.
G.P. is a special series of form $a,ar,a{{r}^{2}},a{{r}^{3}},........,a{{r}^{n-1}}$ where a is first term of G.P and r is common ratio between two consecutive terms.
${{n}^{th}}$ term of G.P. is determined by ${{a}_{n}}=a{{r}^{n-1}}$ and summation of n terms of G.P is $\dfrac{a(1-{{r}^{n}})}{(1-r)}$ and also if a, b, c are in G.P and r is common ratio of an G.P then, $\dfrac{b}{a}=\dfrac{c}{b}=r$ .
Now, in question it is given that sum of two numbers is 6 times their geometric mean.
So, let two numbers be a and b, then
$a+b=6\sqrt{ab}$ …. ( i )
Squaring both sides, we get
${{(a+b)}^{2}}={{\left( 6\sqrt{ab} \right)}^{2}}$
On solving, we get
${{(a+b)}^{2}}=36ab$
Now, subtracting $4ab$ on both sides, we get
${{(a+b)}^{2}}-4ab=36ab-4ab$
${{a}^{2}}+{{b}^{2}}+2ab-4ab=32ab$
${{a}^{2}}+{{b}^{2}}-2ab=32ab$
As, ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$,
So, ${{(a-b)}^{2}}=32ab$
Taking root on both sides, we get
$a-b=\sqrt{32ab}$
$a-b=4\sqrt{2ab}$….. ( ii )
Now, we will use componendo and divdendo method for further calculation.
Let us see what this method does
If $\dfrac{x}{y}=\dfrac{a}{b}$
Then on applying componendo and dividend, expression becomes,
$\dfrac{x+y}{x-y}=\dfrac{a+b}{a-b}$
Now, dividing equation ( i ) by ( ii ), we get
$\dfrac{a+b}{a-b}=\dfrac{6\sqrt{ab}}{4\sqrt{2ab}}$
On simplifying, we get
$\dfrac{a+b}{a-b}=\dfrac{3}{2\sqrt{2}}$…. ( iii )
Applying, componendo and dividendo, we get
\[\dfrac{(a+b)+(a-b)}{(a+b)-(a-b)}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\]
On simplifying, we get
\[\dfrac{2a}{2b}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\]
Or, \[\dfrac{a}{b}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}\]
Hence, numbers a and b, are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$ .

Note: To solve such questions always remember what is A.P and G.P . Always remember the concept of componendo and dividendo that is, If we have $\dfrac{x}{y}=\dfrac{a}{b}$ , then on applying componendo and dividend, expression becomes, $\dfrac{x+y}{x-y}=\dfrac{a+b}{a-b}$.Try to avoid calculation mistakes while solving question.