A) \[\frac{1}{4}-\frac{1}{3\sqrt{2}}\]
B) \[\frac{1}{4}-\frac{1}{4\sqrt{2}}\]
C) \[\frac{1}{4}-\frac{1}{\sqrt{2}}\]
D) \[\frac{1}{4}-\frac{1}{2\sqrt{2}}\]
Correct Answer: D
Solution :
If \[a,b,c\] are in A.P. then \[a+c=2b\] Given \[a+b+c=\frac{3}{4}.......(1)\] \[2b+b=\frac{3}{4}\Rightarrow b=\frac{1}{4}\] \[b=\frac{1}{4}\] If \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in A.P. then \[{{({{b}^{2}})}^{2}}={{a}^{2}}{{c}^{2}}\Rightarrow ac=\pm \frac{1}{16}..........(2)\] From (1) and (2) \[a\pm \frac{1}{16a}=\frac{1}{2}\] \[16{{a}^{2}}-8a\pm 1=0\] If \[16{{a}^{2}}-8a+1=0\Rightarrow a=\frac{1}{4}\] but it is not true; because \[a<b\] If \[16{{a}^{2}}-8a-1=0\Rightarrow a=\frac{8\pm \sqrt{128}}{32}\] \[\Rightarrow a=\frac{1}{4}\pm \frac{1}{2\sqrt{2}}\] But it is given, that\[a<b\] \[\therefore a=\frac{1}{4}-\frac{1}{2\sqrt{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
