The nth term of the geometric progression $- 3, 6, - 12, 24....$ is

- 3 (- 2)^{n + 1}

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3 (- 2)^{n - 1}

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- 3 (2)^{n - 1}

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- 3 (- 2)^{n - 1}

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Solution

The correct option is C $- 3 (- 2)^{n - 1}$
Given series is $- 3, 6, - 12, 24, . . . .$
Here first term $, a = - 3$ , common ratio $, r = - 2$
We know the formula for $n^{t h}$ term,
$a_{n} = a r^{n - 1}$
Thus $a_{n} = - 3 (- 2)^{n - 1}$

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Similar questions

{lim}_{n \to \infty} (\frac{n}{(n + 1) \sqrt{2 n + 1}} + \frac{n}{(n + 2) \sqrt{2 (2 n + 2)}} + \frac{n}{(n + 3) \sqrt{3 (2 n + 3)}} . . . . . + \frac{1}{2 n \sqrt{3}})

\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . . . . . . . . . . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}

$\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}$

If the sum of n terms of the following series:

$\frac{2 n + 1}{2 n - 1} + 3 {(\frac{2 n + 1}{2 n - 1})}^{2} + 5 {(\frac{2 n + 1}{2 n - 1})}^{3} + \dots$ is 36, then value of n is

\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}

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