1.

Which term of the progression 18, -12, 8, … is 512/729, n = ?

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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