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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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Answer» 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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