Question

The product of three numbers of a GP is $\frac{64}{27}$. If the sum of their products when taken in pairs is $\frac{148}{27}$, then find the sum of the three numbers.

• A. $\frac{16}{9}$
• B. $\frac{37}{9}$
• C. $\frac{31}{9}$
• D. $\frac{26}{9}$

Answer 1

Answer: (B)

$\frac{37}{9}$

Let the numbers be $\frac{a}{r},a,ar$ where $r$ is the common ratio

Given, $\frac{a}{r}\times a\times ar=\frac{64}{27}$

$=>a^3=\frac{64}{27}$

$=>a=\frac{4}{3}$

Also, $\frac{a}{r}\times a+a\times ar+\frac{a}{r}\times ar=\frac{148}{27}$

$=>a^2\left(\frac{1}{r}+r+1\right)=\frac{148}{27}$

$=>{\left(\frac{4}{3}\right)}^2\times \left(\frac{1}{r}+r+1\right)=\frac{148}{27}$

$\frac{1}{r}+r+1=\frac{37}{12}$

$=>\frac{1}{r}+r=\frac{25}{12}$

$=>\frac{1+r^2}{r}=\frac{25}{12}$

$=>12+12r^2=25r$

$=>12r^2-25r+12=0$

Solving we get $r=\frac{4}{3};\frac{3}{4}$

We can take any of these values as we will get the same numbers using them

$\frac{a}{r}=\frac{4}{3}\times \frac{4}{3}=\frac{16}{9}$

$a=\frac{4}{3}$

$ar=\frac{4}{3}\times \frac{3}{4}=1$

Their sum $=\frac{16}{9}+\frac{4}{3}+1=\frac{37}{9}$