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{[d(1)=8],[d(n)=d(n-1)*(-5)]:} What is the 3^("rd ") term in the sequence?

{d(1)=8d(n)=d(n−1)⋅(−5) \left\{\begin{array}{l} d(1)=8 \\ d(n)=d(n-1) \cdot(-5) \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?

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Q. {d(1)=8d(n)=d(n−1)⋅(−5) \left\{\begin{array}{l} d(1)=8 \\ d(n)=d(n-1) \cdot(-5) \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?
  1. Understand the sequence definition: Understand the sequence definition.\newlineThe sequence is defined recursively with the first term d(1)d(1) being 88 and each subsequent term d(n)d(n) being the previous term d(n−1)d(n-1) multiplied by −5-5.
  2. Calculate the second term: Calculate the second term using the recursive formula.\newlineWe know that d(1)=8d(1) = 8. To find d(2)d(2), we use the formula d(n)=d(n−1)×(−5)d(n) = d(n-1) \times (-5).\newlineSo, d(2)=d(1)×(−5)=8×(−5)=−40d(2) = d(1) \times (-5) = 8 \times (-5) = -40.
  3. Calculate the third term: Calculate the third term using the recursive formula.\newlineNow that we have d(2)=−40d(2) = -40, we can find d(3)d(3) by using the formula again.\newlined(3)=d(2)×(−5)=−40×(−5)=200d(3) = d(2) \times (-5) = -40 \times (-5) = 200.

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