Find $a_{20}$ of a geometric sequence, if the first few terms of the sequence are given by $- \frac{1}{2}, \frac{1}{4}, - \frac{1}{8}, \frac{1}{16}, \dots$

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Solution

We first find the common ratio $r$ of the GP.
$r = \frac{a_{2}}{a_{1}} = \frac{\frac{1}{4}}{- \frac{1}{2}} = - \frac{1}{2}$
Since the $n^{th}$ term of the GP is given by
$a_{n} = a_{1} r^{n - 1},$ we have
$a_{20} = (- \frac{1}{2}) \times {(- \frac{1}{2})}^{20 - 1} = \frac{1}{2^{20}} .$

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Find a20 of a geometric sequence, if the first few terms of the sequence are given by −12,14,−18,116,⋯

We first find the common ratio r of the GP. r=a2a1=14−12=−12 Since the nth term of the GP is given by an=a1rn−1, we have a20=(−12)×(−12)20−1=1220.

Find $a_{20}$ of a geometric sequence, if the first few terms of the sequence are given by $- \frac{1}{2}, \frac{1}{4}, - \frac{1}{8}, \frac{1}{16}, \dots$

We first find the common ratio $r$ of the GP.
$r = \frac{a_{2}}{a_{1}} = \frac{\frac{1}{4}}{- \frac{1}{2}} = - \frac{1}{2}$
Since the $n^{th}$ term of the GP is given by
$a_{n} = a_{1} r^{n - 1},$ we have
$a_{20} = (- \frac{1}{2}) \times {(- \frac{1}{2})}^{20 - 1} = \frac{1}{2^{20}} .$