Sequence And Series Question 399
Question: Three numbers are in G.P. such that their sum is 38 and their product is 1728. The greatest number among them is:
Options:
A) 18
B) 16
C) 14
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let the required three numbers of G.P. be $ \frac{a}{r}, $ a and ar. Then, their sum $ =\frac{a}{r}+a+ar=38 $
$ \Rightarrow a( \frac{1+r+r^{2}}{r} )=38 $ ?.. (i) product $ =\frac{a}{r}\times a\times ar=1728 $
$ \Rightarrow ,a^{3}={{(12)}^{3}}\therefore a=12 $ ….. (ii) Substitute the value of a, in equation (i), we get
$ \therefore 12\times ( \frac{1+r+r^{2}}{r} )=38 $
$ \Rightarrow ,6+6r+6r^{2}=19r,\Rightarrow ,6r^{2}-13r+6=0 $
$ \Rightarrow (3r-2)(2r-3)=0 $
$ \therefore ,r=\frac{2}{3} $ or $ \frac{3}{2} $
Hence, the required numbers are 18, 12, 8 or 8,
12, 18
$ \therefore $ Greatest number = 18
BETA
