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{[b(1)=15],[b(n)=b(n-1)*(-3)]:} What is the 4^("th ") term in the sequence?

{b(1)=15b(n)=b(n−1)⋅(−3) \left\{\begin{array}{l} b(1)=15 \\ b(n)=b(n-1) \cdot(-3) \end{array}\right. \newlineWhat is the 4th  4^{\text {th }} term in the sequence?

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Q. {b(1)=15b(n)=b(n−1)⋅(−3) \left\{\begin{array}{l} b(1)=15 \\ b(n)=b(n-1) \cdot(-3) \end{array}\right. \newlineWhat is the 4th  4^{\text {th }} term in the sequence?
  1. Given information: We are given the first term of the sequence, b(1)=15b(1) = 15, and a recursive formula for the sequence, b(n)=b(n−1)×(−3)b(n) = b(n-1) \times (-3). To find the 44th term, we need to apply the recursive formula three times starting from the first term.
  2. Find second term: First, we find the second term using the recursive formula. The second term is b(2)=b(1)×(−3)b(2) = b(1) \times (-3).\newlineCalculation: b(2)=15×(−3)=−45b(2) = 15 \times (-3) = -45.
  3. Find third term: Next, we find the third term using the recursive formula. The third term is b(3)=b(2)×(−3)b(3) = b(2) \times (-3).\newlineCalculation: b(3)=−45×(−3)=135b(3) = -45 \times (-3) = 135.
  4. Find fourth term: Finally, we find the fourth term using the recursive formula. The fourth term is b(4)=b(3)×(−3)b(4) = b(3) \times (-3).\newlineCalculation: b(4)=135×(−3)=−405b(4) = 135 \times (-3) = -405.

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