Solve this number series:
113, 114 , 118, ? , 143 , 168

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Solution

Difference is the square of number I.e. 1,4,9,16,25

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Similar questions

1 + \frac{2}{11} + \frac{5}{11^{2}} + \frac{10}{11^{3}} + \frac{17}{11^{4}} + . .

\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}

\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}

\begin{matrix} 143 & (56) & 255 218 & (. . . .) & 114 \end{matrix}

\tan^{- 1} \frac{1}{7} + \tan^{- 1} \frac{1}{13} = \tan^{- 1} \frac{2}{9}

\tan^{- 1} \frac{1}{4} + \tan^{- 1} \frac{2}{9} = \frac{1}{2} \cos^{- 1} \frac{3}{5} = \frac{1}{2} \sin^{- 1} (\frac{4}{5})

\tan^{- 1} \frac{2}{3} = \frac{1}{2} \tan^{- 1} \frac{12}{5}

\tan^{- 1} \frac{1}{7} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{4}

\sin^{- 1} \frac{4}{5} + 2 \tan^{- 1} \frac{1}{3} = \frac{π}{2}

\sin^{- 1} \frac{12}{13} + \cos^{- 1} \frac{4}{5} + \tan^{- 1} \frac{63}{16} = π

2 \sin^{- 1} \frac{3}{5} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{8} = \tan^{- 1} \frac{4}{7}

2 \tan^{- 1} \frac{3}{4} - \tan^{- 1} \frac{17}{31} = \frac{π}{4}

2 \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{7}) = \tan^{- 1} (\frac{31}{17})

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