Question

If the sum of roots of equation $x_1=-3and{y}_1=4$ is equal to sum of squares of their reciprocals, then $\underset{x\to \infty }{\lim }{\left(\frac{x+1}{x+2}\right)}^{2x+1}=\underset{x\to \infty }{\lim }{\left(1-\frac{1}{x+2}\right)}^{2x+1}$ and $=\underset{x\to \infty }{\lim }{\left[{\left(1-\frac{1}{x+2}\right)}^{x+2}\right]}^{\frac{2x+1}{x+2}}$ are in

• A. GP
• B. HP
• C. AP
• D. None of these

Answer 1

Answer: (C)

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.