If the arithmetic mean of a and b is $\frac{{\mathrm{a}}^{\mathrm{n}} + {\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n} - 1} + {\mathrm{b}}^{\mathrm{n} - 1}}$, then the value of n is

• - 1

• 0

• 1

• None of the above

$\frac{{\displaystyle \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + ...}}{1 + {\displaystyle \frac{1}{3!}} + {\displaystyle \frac{1}{5!}} + ...}$ equals

• e + 1

• $\frac{\mathrm{e} - 1}{\mathrm{e} + 1}$

• e - 1

• None of these

The sum of the series

$\left(\frac{\mathrm{a} - \mathrm{b}}{\mathrm{a}}\right) + \frac{1}{2}{\left(\frac{\mathrm{a} - \mathrm{b}}{\mathrm{a}}\right)}^2 + \frac{1}{3}{\left(\frac{\mathrm{a} - \mathrm{b}}{\mathrm{a}}\right)}^3 + ... \infty$ is

• ${\log }_{\mathrm{e}}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)$

• ${\log }_{\mathrm{e}}\left(\frac{\mathrm{a} - \mathrm{b}}{\mathrm{a}}\right)$

• ${\log }_{\mathrm{e}}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$

• None of these

The harmonic mean of the roots of the equation

$\left(5 + \sqrt{2}\right){\mathrm{x}}^2 - \left(4 + \sqrt{5}\right)\mathrm{x} + 8 + 2\sqrt{5} = 0$ is

• 2

• 4

• 6

• 8

The sum of n terms of the following series 1 + (1 + x) + (1 + x + x²) +... will be

• $\frac{1 - {\mathrm{x}}^{\mathrm{n}}}{1 - \mathrm{x}}$

• $\frac{\mathrm{x}\left(1 - {\mathrm{x}}^{\mathrm{n}}\right)}{1 - \mathrm{x}}$

• $\frac{\mathrm{n}\left(1 - \mathrm{x}\right) - \mathrm{x}\left(1 - {\mathrm{x}}^{\mathrm{n}}\right)}{{\left(1 - \mathrm{x}\right)}^2}$

• None of the above

116.

If A₁, A_2; G₁, G₂ and H₁, H₂ be two AM's, GM's and HM's between two quantities, then the value of $\frac{{\mathrm{G}}_1{\mathrm{G}}_2}{{\mathrm{H}}_1{\mathrm{H}}_2}$ is

• $\frac{{\mathrm{A}}_1 + {\mathrm{A}}_2}{{\mathrm{H}}_1 + {\mathrm{H}}_2}$

• $\frac{{\mathrm{A}}_1 - {\mathrm{A}}_2}{{\mathrm{H}}_1 + {\mathrm{H}}_2}$

• $\frac{{\mathrm{A}}_1 + {\mathrm{A}}_2}{{\mathrm{H}}_1 - {\mathrm{H}}_2}$

• $\frac{{\mathrm{A}}_1 - {\mathrm{A}}_2}{{\mathrm{H}}_1 - {\mathrm{H}}_2}$

1² + 1 + 2² + 2 + 3² + 3 + ... + n² + n is equal to

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)}{2}$

• ${\left[\frac{\mathrm{n}\left(\mathrm{n} + 1\right)}{2}\right]}^2$

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)\left(\mathrm{n} + 2\right)}{3}$

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)\left(\mathrm{n} + 2\right)\left(\mathrm{n} + 3\right)}{4}$

2^1/4 . 4^1/8 . 8^1/16 . 16^1/32 ... is equal to

• 1

• 2

• $\frac{3}{2}$

• $\frac{5}{2}$

The sum of the series

$\sum {\left(- 1\right)}^{\mathrm{r}}{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}\left(\frac{1}{2^{\mathrm{r}}} + \frac{3^{\mathrm{r}}}{2^{2\mathrm{r}}} + \frac{7^{\mathrm{r}}}{2^{3\mathrm{r}}} + \frac{{15}^{\mathrm{r}}}{2^{4\mathrm{r}}} + ... \mathrm{m} \mathrm{terms}\right)$ is

• $\frac{2^{\mathrm{mn}} - 1}{2^{\mathrm{mn}}\left(2^{\mathrm{n}} - 1\right)}$

• $\frac{2^{\mathrm{mn}} - 1}{2^{\mathrm{n}} - 1}$

• $\frac{2^{\mathrm{mn}} + 1}{2^{\mathrm{n}} + 1}$

• None of these

If n = (1999)!, Then $\sum _{\mathrm{x} = 1}^{1999}{\log }_{\mathrm{n}}\left(\mathrm{x}\right)$ is equal to

• 1

• 0

• $\sqrt[1999]{1999}$

• - 1