A) \[\frac{{{A}_{1}}+{{A}_{2}}}{{{H}_{1}}+{{H}_{2}}}\]
B) \[\frac{{{A}_{1}}-{{A}_{2}}}{{{H}_{1}}+{{H}_{2}}}\]
C) \[\frac{{{A}_{1}}+{{A}_{2}}}{{{H}_{1}}-{{H}_{2}}}\]
D) \[\frac{{{A}_{1}}-{{A}_{2}}}{{{H}_{1}}-{{H}_{2}}}\]
Correct Answer: A
Solution :
Let the two quantities be \[a\] and \[b\]. Then \[a,\ {{A}_{1}},\ {{A}_{2}},\ b\] are in A.P. \ \[{{A}_{1}}-a=b-{{A}_{2}}\Rightarrow {{A}_{1}}+{{A}_{2}}=a+b\] ......(i) Again \[a,\ {{G}_{1}},\ {{G}_{2}},\ b\] are in G.P. \[\therefore \]\[\frac{{{G}_{1}}}{a}=\frac{b}{{{G}_{2}}}\Rightarrow {{G}_{1}}{{G}_{2}}=ab\] .....(ii) Also \[a,\ {{H}_{1}},\ {{H}_{2}},\ b\] are in H.P. \[\therefore \]\[\frac{1}{{{H}_{1}}}-\frac{1}{a}=\frac{1}{b}-\frac{1}{{{H}_{2}}}\Rightarrow \frac{1}{{{H}_{1}}}+\frac{1}{{{H}_{2}}}=\frac{1}{a}+\frac{1}{b}\] \[\Rightarrow \]\[\frac{{{H}_{1}}+{{H}_{2}}}{{{H}_{1}}{{H}_{2}}}=\frac{a+b}{ab}=\frac{{{A}_{1}}+{{A}_{2}}}{{{G}_{1}}{{G}_{2}}}\] [By (i) and (ii)] \[\therefore \]\[\frac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\frac{{{A}_{1}}+{{A}_{2}}}{{{H}_{1}}+{{H}_{2}}}\].
You need to login to perform this action.
You will be redirected in
3 sec
