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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 :


∴ 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 \(\frac{1}{3^{n-1}}\)


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




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

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