A) \[\frac{a+b}{ab}\]
B) \[\frac{a+b}{2ab}\]
C) \[\frac{2ab}{a+b}\]
D) \[\frac{ab}{a+b}\]
Correct Answer: A
Solution :
Given that \[a,\ {{A}_{1}},\ {{A}_{2}},\ b\] are in A.P. Therefore \[{{A}_{1}}=\frac{a+{{A}_{2}}}{2},\ {{A}_{2}}=\frac{{{A}_{1}}+b}{2}\] \[\Rightarrow \] \[x+y+z=15\] \[\Rightarrow \] \[=9+15+a=\frac{5}{2}(9+2)\] or \[{{A}_{1}}+{{A}_{2}}=a+b\] ?..(i) and \[a,\ {{G}_{1}},\ {{G}_{2}},\ b\] are in G.P. Therefore \[G_{1}^{2}=a{{G}_{2}},\ G_{2}^{2}=b{{G}_{1}}\] ?..(ii) \[\Rightarrow G_{1}^{2}G_{2}^{2}=ab{{G}_{1}}{{G}_{2}}\Rightarrow {{G}_{1}}{{G}_{2}}=ab\] Hence \[\frac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}=\frac{a+b}{ab}\] Trick: Let\[a=1,\ b=2\], then \[{{A}_{1}}+{{A}_{2}}=1+2=3\] and \[{{G}_{1}}\ .\ {{G}_{2}}=2\times 1=2\] \[\therefore \ \]\[\frac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}=\frac{3}{2}\], which is given by (a).
You need to login to perform this action.
You will be redirected in
3 sec
