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

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Full solution

Q. egin{cases}b(11)=−54-54\ b(n)=b(n−1-1)\cdot \dfrac{44}{33} \end{cases}. what is the 44th term in the sequence?
  1. Given terms: We are given the first term of the sequence b(1)=−54 b(1)=-54 and a recursive formula for the sequence b(n)=b(n−1)⋅43 b(n)=b(n-1)\cdot \dfrac{4}{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, let's find the second term b(2) b(2) using the recursive formula:\newlineb(2)=b(1)⋅43 b(2) = b(1) \cdot \dfrac{4}{3} \newlineb(2)=−54⋅43 b(2) = -54 \cdot \dfrac{4}{3} \newlineb(2)=−72 b(2) = -72
  3. Find third term: Next, we find the third term b(3) b(3) using the second term:\newlineb(3)=b(2)⋅43 b(3) = b(2) \cdot \dfrac{4}{3} \newlineb(3)=−72⋅43 b(3) = -72 \cdot \dfrac{4}{3} \newlineb(3)=−96 b(3) = -96
  4. Find fourth term: Finally, we find the fourth term b(4) b(4) using the third term:\newlineb(4)=b(3)⋅43 b(4) = b(3) \cdot \dfrac{4}{3} \newlineb(4)=−96⋅43 b(4) = -96 \cdot \dfrac{4}{3} \newlineb(4)=−128 b(4) = -128

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