$$\begin{gathered} \mathrm{If} \mathrm{S} \mathrm{is} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{first} 10 \mathrm{terms} \mathrm{of} \mathrm{the} \mathrm{series} {\tan }^{-1}\left(\frac{1}{3}\right) + {\tan }^{-1}\left(\frac{1}{7}\right) + {\tan }^{-1}\left(\frac{1}{13}\right) + {\tan }^{-1}\left(\frac{1}{3}\right) \\ + {\tan }^{-1}\left(\frac{1}{21}\right) + ... , \mathrm{then} \tan \left(\mathrm{S}\right) = ? \end{gathered}$$

• $- \frac{6}{5}$

• $\frac{5}{11}$

• $\frac{5}{6}$

• $\frac{10}{11}$

$\mathrm{If} 2^{10} + 2^9 . 3^1 +\hspace{0.17em}{2}^8 . 3^2 +\hspace{0.17em}... + 2 . 3^9 +3^{10} = 3^{10}= \mathrm{S} - 2^{11} \mathrm{Then} \mathrm{S} = ?$

• $3^{11}$

• $2 . 3^{11}$

• $\frac{3^{11}}{2} + 2^{10}$

• $3^{11} - 2^{11}$

If the sum of the second,third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

• $\frac{2}{13}\left(3^{50} - 1\right)$

• $\frac{1}{26}\left(3^{49} - 1\right)$

• $\frac{1}{13}\left(3^{50} - 1\right)$

• $\frac{1}{26}\left(3^{50} - 1\right)$

If the sum of the first 20 terms of the series log$\left({\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}\left(\mathrm{x}\right)\right) + \left({\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}\left(\mathrm{x}\right)\right) + \left({\log }_{7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\left(\mathrm{x}\right)\right) + ... \mathrm{is} 460$,

then x = ?

• e²

• $7^{\raisebox{1ex}{$46$}\!\left/ \!\raisebox{-1ex}{$21$}\right.}$

• 7²

• $7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

235.

The coefficient of x⁴ in the expansion of (1 + x + x² + x³)⁶ in powers of x, is .

Out of 11 consecutive natural number if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference is :

• $\frac{10}{99}$

• $\frac{15}{101}$

• $\frac{5}{33}$

• $\frac{5}{101}$

Let a , b, c, d and ay non zero distinct real numbers such that (a² + b² + c²)p² – 2(ab + bc + cd)p + (b² + c² + d²) = 0. Then :

• a, b, c, d are in A.P.

• a, c, p are in G.P.

• a, b, c, d are in G.P.

• a, c, p, are in A.P.

The 100th term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... , is

• 12

• 13

• 14

• 15

Let S_n denotes the sum of first n terms of an AP.If S₄ = - 34, S₅ = - 60 and S₆ = - 93, then the common difference and the first term of the AP are respectively

• - 7, 2

• 7, - 4

• 7, - 2

• - 7, - 2

An AP has the property that the sum of first ten terms is half the sum of next ten terms. If the second term is 13, then the common difference is

• 3

• 2

• 5

• 4