Which term of the progression 18, -12, 8, … is \( \dfrac{512}{729}\) ?

Asked by Sakshi | 1 year ago |  65

1 Answer

Solution :-

By using the formula,

Tn = arn-1

a = 18

r = \( \dfrac{-12}{18}\)

\( \dfrac{-2}{3}\)

Tn =\( \dfrac{512}{729}\)

n = ?

Tn = arn-1

8 = n – 1

n = 8 + 1

= 9

9th term of the Progression is \( \dfrac{512}{729}\)

Answered by Sakshi | 1 year ago

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