A) \[a=-\frac{1}{4},b=1\]
B) \[a=\frac{1}{12},b=\frac{1}{9}\]
C) (a) and (b) both are true
D) None of these
Correct Answer: C
Solution :
If \[\frac{1}{16},\,\,a,b\]are in G.P., then \[{{a}^{2}}=\frac{b}{16}\] or \[16\,{{a}^{2}}=b\] .....(i) and if \[a,b,\frac{1}{6}\]are in H.P., then \[b=\frac{2.a.\frac{1}{6}}{a+\frac{1}{6}}=\frac{2a}{6a+1}\] ?..(ii) From (i) and (ii), \[16{{a}^{2}}=\frac{2a}{6a+1}\] or \[2a\left( 8a-\frac{1}{6a+1} \right)=0\] or \[8a\,(6a+1)-1=0\] or \[48{{a}^{2}}+8a-1=0\], \[(\because a\ne 0)\] or \[(4a+1)(12a-1)=0\] \[\therefore \] \[a=-\frac{1}{4},\frac{1}{12}\] When \[a=-\frac{1}{4}\] then from (i), \[b=16\,{{\left( -\frac{1}{4} \right)}^{2}}=1\] When \[a=\frac{1}{12}\] then from (i), \[b=16{{\left( \frac{1}{12} \right)}^{2}}=\frac{1}{9}\] Therefore, \[a=-\frac{1}{4},b=1\] or \[a=\frac{1}{12},b=\frac{1}{9}\].
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