Question

10. Find the common ratio of three numbers in G.P whose product is 216 and the sum of the products taken in pairs is 114.
4 or 14
2/3 or 3/2
2 or 1/2
O %4 or 413​

Answer 1

Answer:

The required answer is $4,6,9$ or $9,6,4$

Step-by-step explanation:

Given: Product=216, sum of product in pair=156

To find: Three number.

Solutions:

Let $3$ numbers in GP be $a, a/r, ar$

Now,

$\frac{\mathrm{a}}{\mathrm{r}} \cdot \mathrm{a} \cdot \mathrm{ar}=216$

Calculate the value of $a$.

$a^{3}=216$

$a=6$

Calculate the  three numbers in G.P.

$\frac{\mathrm{a}}{\mathrm{r}} \cdot \mathrm{a} / \mathrm{ar}+\mathrm{ar} \cdot \frac{\mathrm{a}}{\mathrm{r}}=114$

$\frac{36}{r}+36 r+36=114$

$\frac{36}{r}+36 r=114-36$

$\frac{36}{r}+36 \mathrm{r}=78$

$36\left[\frac{1}{r}+r\right]=78$

$36\left[1+\frac{r^{2}}{r}\right]=78$

$36 r^{2}-78 r+36=0$

Common factor.

$&36 r^{2}-78 r+36=0$

$&6\left(6 r^{2}-13 r+6\right)=0$

By the same factor, divide both sides.

$&6\left(6 r^{2}-13 r+6\right)=0$

$&6 r^{2}-13 r+6=0$

Use the quadratic formula

$r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Identify $a, b$ and $c$ from the original equation once it is in standard form, and then enter them into the quadratic formula.

$&6 r^{2}-13 r+6=0$

$&a=6$

$&b=-13$

$c=6$

$r=\frac{-(-13) \pm \sqrt{(-13)^{2}-4 \cdot 6 \cdot 6}}{2 \cdot 6}$

Simplify.

$r=\frac{13 \pm 5}{12}$

Separate the equations.

Divide the equation into two halves, one with a plus and the other with a minus, to find the value of the unknown variable.

$r=\frac{13+5}{12}$

$r=\frac{13-5}{12}$

To find each solution, rearrange and isolate the variable.

$&r=\frac{3}{2}$ or $&r=\frac{2}{3}$

$&\text { if } \mathrm{r}=\frac{3}{2} \& \mathrm{a}=6$

$&\text { then, } \mathrm{a}_{1}=\frac{\mathrm{a}}{\mathrm{r}}$

$&=\frac{6}{\frac{3}{2}}$

$&=4$

$&\mathrm{a}_{2}=\mathrm{a}=6$

$&\mathrm{a}_{3}=\mathrm{ar}$

$&=6 \times \frac{3}{2}$

$&=9$

$&\text { if } \mathrm{r}=\frac{2}{3} \& \mathrm{a}=6$

$&\text { then, }$

$&\mathrm{a}_{1}=\frac{\mathrm{a}}{\mathrm{r}}$

$&=\frac{6}{\frac{2}{3}}$

$&=9$

$&\mathrm{a}_{2}=\mathrm{a}=6$

$&\mathrm{a}_{3}=\mathrm{ar}=4$

if $&r=\frac{3}{2}$, three numbers are $4,6,9$.

If $&r=\frac{2}{3}$, three numbers are $9,6,4$.

Hence, the three numbers in G.P. are $4,6,9$ or $9,6,4$.

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Answer 2

Answer:

The terms of the GP are either 4, 6, and 9 or 9, 6, and 4, with the common ratio to be either $\frac{2}{3} \; or \; \frac{3}{2}$

Step-by-step explanation:

Given - the product of 3 numbers in GP is 216, and the sum of the products when in pairs is 114.

To find - the 3 numbers in GP

Solution -

Let, the 3 numbers in GP to find be $\frac{a}{r} , a, ar$.

We are given that the product of the 3 numbers is 216, and we can solve that as follows.

$\frac{a}{r} \times a \times ar = 216\\\\\implies a^3 = 216\\\implies a = 6$

We are also given that the product of the 3 numbers when taken in pairs is 114, which we will solve as follows -

$(\frac{a}{r} \times a) + (a \times ar) + (ar \times \frac{a}{r}) = 114 \\\\(\frac{36}{r}) + (36r) + (36) = 114\\\\(\frac{36}{r}) + (36r) = 78\\\\36 [1+ (\frac{r^2}{r})] = 78\\\\36r^2 - 78r + 36 = 0\\6(6r^2-13r+6) = 0\\6r^2 - 13r + 6 = 0$

By solving the given equation using quadratic equations, we get $r = \frac{3}{2} \; or \; r = \frac{2}{3}$

Using these solutions, we can find the numbers to be 4, 6, and 9 or 9, 6, and 4.

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