Question

Three numbers are in G.P. such that their sum is $38$ and their product is $1728$.

The greatest number among them is

• A. $18$
• B. $16$
• C. $14$
• D. None of these

Answer 1

The correct option is A

$18$

Explanation for the correct answer:

Let the three consecutive numbers in an arithmetic progression be

$\frac{a}{r},a,ar$

It is given that the product of these numbers is $1728$

$\Rightarrow$ $\frac{a}{r}\times a\times ar=1728$

$\Rightarrow$ $a^3=1728$

Taking cube root on both sides we get

$\Rightarrow$ $a=12$

It is given that the sum of these numbers is $38$

$\Rightarrow$ $\frac{a}{r}+a+ar=38$

Substituting the value of $a=12$ we get

$\Rightarrow$ $\frac{12}{r}+12+12r=38$

$\Rightarrow$ $12\left(\frac{1}{r}+r\right)=26$

$\Rightarrow$ $12\left(r^2+1\right)=26r$

$\Rightarrow$ $6r^2-13r+6=0$

$\Rightarrow$ $6r^2-9r-4r+6=0$

$\Rightarrow$ $3r\left(2r-3\right)-2\left(2r-3\right)=0$

$\Rightarrow$ $\left(3r-2\right)(2r-3)=0$

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

Hence, the terms in the geometric progression are

$\frac{a}{r}=\frac{12}{{\displaystyle \frac{2}{3}}}=18$ or $\frac{a}{r}=\frac{12}{{\displaystyle \frac{3}{2}}}=8$

$ar=12\times \frac{2}{3}=8$ or $ar=12\times \frac{3}{2}=18$

Hence, the geometric progression is $18,12,8$ or $8,12,18$.

Hence, the greatest number of the geometric progression is $18$.

Hence, option (A) is the correct answer.