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egin{equation}\begin{cases}b(11)=−54-54\ b(n)=b(n−1-1)\cdot \dfrac{44}{33} \end{cases}\end{equation} what is the fourth term in the sequence?

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Full solution

Q. egin{equation}\begin{cases}b(11)=−54-54\ b(n)=b(n−1-1)\cdot \dfrac{44}{33} \end{cases}\end{equation} what is the fourth term in the sequence?
  1. First Term Given: To find the fourth term in the sequence, we need to apply the recursive formula three times starting from the first term.\newlineFirst term b(1) b(1) is given as −54-54.
  2. Calculate Second Term: Now we find the second term b(2) b(2) using the recursive formula b(n)=b(n−1)⋅43 b(n)=b(n-1)\cdot \dfrac{4}{3} .\newlineSo, b(2)=b(1)⋅43=−54⋅43 b(2) = b(1) \cdot \dfrac{4}{3} = -54 \cdot \dfrac{4}{3} .\newlineCalculating this gives us b(2)=−72 b(2) = -72 .
  3. Calculate Third Term: Next, we find the third term b(3) b(3) using the same recursive formula.\newlineb(3)=b(2)⋅43=−72⋅43 b(3) = b(2) \cdot \dfrac{4}{3} = -72 \cdot \dfrac{4}{3} .\newlineCalculating this gives us b(3)=−96 b(3) = -96 .
  4. Calculate Fourth Term: Finally, we find the fourth term b(4) b(4) using the recursive formula.\newlineb(4)=b(3)⋅43=−96⋅43 b(4) = b(3) \cdot \dfrac{4}{3} = -96 \cdot \dfrac{4}{3} .\newlineCalculating this gives us b(4)=−128 b(4) = -128 .

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