What is the $8^{t h}$ term of the harmonic sequence whose nth term is $a_{n} = \frac{2}{n^{2} + 1}$ ?

\frac{1}{65}

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\frac{2}{65}

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\frac{3}{65}

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\frac{4}{65}

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Solution

The correct option is A $\frac{2}{65}$
Given, $a_{n} = \frac{2}{n^{2} + 1}$
Substituting $n - 8$ in the given equation
Therefore, $a_{8} = \frac{2}{8^{2} + 1}$
$\Rightarrow a_{8} = \frac{2}{65}$

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Similar questions

8^{t h}

15^{t h}

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19^{t h}

65

The 10th term of an arithmetic sequence is 65 and its 15th term is 20. Then 20th term is

\frac{3}{2} + \frac{9}{4} + \frac{25}{8} + \frac{65}{16} + . . .

$c o s \frac{π}{65} c o s \frac{2 π}{65} c o s \frac{4 π}{65} c o s \frac{8 π}{65} c o s \frac{16 π}{65} c o s \frac{32 π}{65} = \frac{1}{64}$

1 + 5 + 19 + 65 + . . . . . .

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