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Which term of the progression 18, -12, 8, … is 512/729, n = ? |
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Answer» Tn = arn-1 a = 18, r = \(\frac{-12}{18}\) = \(\frac{-2}{3}\), Tn = \(\frac{512}{729}, n = ?\) \(\therefore\)\(\frac{512}{729}\) = 18 x (\(\frac{-2}{3}\))n-1 \(\Rightarrow\) \(\frac{512}{729}\) = 18 x (\(\frac{-2}{3}\))n x \(\frac{-3}{2}\) \(\Rightarrow\)\(\frac{512}{19683}\) = \((\frac{2}{3})^3\) \(\Rightarrow\) n = 9 |
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