If the sum of three numbers in G.P is 7/64 and the product of the extremes is 1/1024 then the numbers are

Correct option is (C) \(\frac{1}{64} ,\frac{1}{32} , \frac{1}{16}\)

Let required three numbers in G.P. are \(\frac ar,a,ar.\)

\(\therefore\) Product of extremes \(=\frac ar.ar=a^2\)

\(\therefore a^2=\frac1{1024}=(\frac1{32})^2\)

\(a=\frac1{32}\)

Also given that sum of these numbers are \(\frac{7}{64}.\)

i.e., \(\frac ar+a+ar=\frac7{64}\)

\(\Rightarrow\frac1{32r}+\frac1{32}+\frac r{32}=\frac7{64}\)

\(\Rightarrow\frac2r+2+2r=7\)      (Multiply both sides by 64)

\(\Rightarrow2r^2+2r+2=7r\)    (Multiply both sides by r)

\(\Rightarrow2r^2-5r+2=0\)

\(\Rightarrow2r^2-4r-r+2=0\)

\(\Rightarrow2r(r-2)-1(r-2)=0\)

\(\Rightarrow(r-2)(2r-1)=0\)

\(\Rightarrow r-2=0\;or\;2r-1=0\)

\(\Rightarrow r=2\;or\;r=\frac12\)

\(\therefore\) Required number are either \(\frac{a}{r}=\frac{1}{32}\times\frac{1}{2}=\frac{1}{64},\) \(a=\frac{1}{32},ar=\frac{1}{32}\times2=\frac{1}{16}\)

or \(\frac{a}{r}=\frac{1}{32}\times2=\frac{1}{16},a=\frac{1}{32},\) \(ar=\frac{1}{32}\times\frac12=\frac{1}{64}\)

Hence, required numbers are \(\frac{1}{64} ,\frac{1}{32} , \frac{1}{16}.\)

Correct option is C) 1/64 , 1/32 , 1/16