A) 18
B) 16
C) 14
D) None of these
Correct Answer: A
Solution :
[a] Let the required three numbers of G.P. be |
\[\frac{a}{r},\] a and ar. |
Then, their sum \[=\frac{a}{r}+a+ar=38\] |
\[\Rightarrow \,\,\,\,\,\,a\left( \frac{1+r+{{r}^{2}}}{r} \right)=38\] ?.. (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 |
\[\therefore \,\,\,\,\,\,12\times \left( \frac{1+r+{{r}^{2}}}{r} \right)=38\] |
\[\Rightarrow \,\,\,\,\,6+6r+6{{r}^{2}}=19r\,\,\,\Rightarrow \,\,\,6{{r}^{2}}-13r+6=0\] |
\[\Rightarrow \,\,\,\,\,\,(3r-2)\,\,(2r-3)=0\] \[\therefore \,\,\,\,\,r=\frac{2}{3}\] or \[\frac{3}{2}\] |
Hence, the required numbers are 18, 12, 8 or 8, |
12, 18 |
\[\therefore \] Greatest number = 18 |
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