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{[b(1)=-500],[b(n)=b(n-1)*(4)/(5)]:} What is the 3^("rd ") term in the sequence?

{b(1)=−500b(n)=b(n−1)⋅45 \left\{\begin{array}{l} b(1)=-500 \\ b(n)=b(n-1) \cdot \frac{4}{5} \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?

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Full solution

Q. {b(1)=−500b(n)=b(n−1)⋅45 \left\{\begin{array}{l} b(1)=-500 \\ b(n)=b(n-1) \cdot \frac{4}{5} \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?
  1. Given information: We are given the first term of the sequence, b(1)=−500b(1) = -500, and a recursive formula for the sequence, b(n)=b(n−1)×(45)b(n) = b(n-1) \times \left(\frac{4}{5}\right). To find the third term, b(3)b(3), we need to first find the second term, b(2)b(2), using the first term and the recursive formula.
  2. Calculate second term: Using the recursive formula, we calculate the second term as follows:\newlineb(2)=b(1)×(45)=−500×(45)=−400.b(2) = b(1) \times \left(\frac{4}{5}\right) = -500 \times \left(\frac{4}{5}\right) = -400.\newlineWe have found the second term of the sequence without any mathematical errors.
  3. Find third term: Now, we use the second term to find the third term using the same recursive formula:\newlineb(3)=b(2)×(45)=−400×(45)=−320.b(3) = b(2) \times \left(\frac{4}{5}\right) = -400 \times \left(\frac{4}{5}\right) = -320.\newlineWe have found the third term of the sequence without any mathematical errors.

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