A) GP
B) HP
C) AP
D) None of these
Correct Answer: C
Solution :
Let roots of equation \[{{Q}_{1}}{{R}_{2}}\ne {{Q}_{2}}{{R}_{1}}\]are\[{{Q}_{1}}{{R}_{2}}={{Q}_{2}}{{R}_{1}}\] and\[s=\frac{{{t}^{2}}}{4}\] Then, \[T\propto V\]and \[T\propto {{V}^{2}}\] According to the question, \[T\propto \frac{1}{{{V}^{2}}}\] \[T\propto \frac{1}{V}\] \[\text{6}\times \text{1}{{0}^{-\text{7}}}\text{A}-{{\text{m}}^{\text{2}}}\] \[\text{5 g}/\text{c}{{\text{m}}^{\text{3}}}\] \[\text{8}\text{.3}\times \text{1}{{0}^{\text{6}}}\] \[\text{1}.\text{2}\times \text{1}{{0}^{-\text{7}}}\] \[\text{3}\times \text{1}{{0}^{-\text{6}}}\] \[CaC{{l}_{2}}\] \[\text{MgS}{{\text{O}}_{\text{4}}}\] \[\text{MgS}{{\text{O}}_{\text{4}}}\] \[CaC{{l}_{2}}\] Hence, \[A\to B,B\to C\] and \[C\to A\] are in AP.You need to login to perform this action.
You will be redirected in
3 sec