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Find the nth term and sum up to 13terms of the sequence: 3,−1, 1/3, -1/9...

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Given sequence is :

3,-1,\(\frac{1}{3}\),\(\frac{-1}{9}\),....

∴ a1 = 3, a2 = -1, a3\(\frac{1}{3}\), a4\(\frac{-1}{9}\),....

Common ration = r = \(\frac{a_4}{a_3}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_2}{a_1}\) = \(\frac{-1}{3}\) < 1

nth term of given sequence is,

an = arn-1

= 3 x (\(\frac{1}{3}\))n-1

= (-1)n-1 x 3 x \(\frac{1}{3^{n-1}}\)

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

Sum upto 13 terms = S13\(\frac{a(1-r^{13})}{1-r}\)

\(\frac{3(1-(\frac{-1}{3})^{13})}{1-(\frac{-1}{3})}\)

\(\frac{3(1+\frac{1}{3^{13}})}{1+\frac{1}{3}}\)

\(\frac{3(1+\frac{1}{3^{13}})}{\frac{4}{3}}\)

\(\frac{9}{4}\)(1 + \(\frac{1}{3^{13}}\))

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