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 :
18
Solution :
Let the required three numbers of G.P. be $$\frac{a}{r},a{\text{ and }}ar.$$
Then, their sum $$ = \frac{a}{r} + a + ar = 38$$
$$\eqalign{
& \Rightarrow a\left( {\frac{{1 + r + {r^2}}}{r}} \right) = 38\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{product}} = \frac{a}{r} \times a \times ar = 1728 \cr
& \Rightarrow {a^3} = {\left( {12} \right)^3} \cr
& \therefore a = 12\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Substitute the value of $$a,$$ in equation (i), we get
$$\eqalign{
& \therefore 12 \times \left( {\frac{{1 + r + {r^2}}}{r}} \right) = 38 \cr
& \Rightarrow 6 + 6r + 6{r^2} = 19r \cr
& \Rightarrow 6{r^2} - 13r + 6 = 0 \cr
& \Rightarrow \left( {3r - 2} \right)\left( {2r - 3} \right) = 0 \cr
& \therefore r = \frac{2}{3}{\text{ or }}\frac{3}{2} \cr} $$
Hence, the required numbers are 18, 12, 8 or 8, 12, 18
∴ Greatest number = 18