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{[b(1)=-2],[b(n)=b(n-1)-7]:} Find the 3^("rd ") term in the sequence.

{b(1)=−2b(n)=b(n−1)−7 \left\{\begin{array}{l} b(1)=-2 \\ b(n)=b(n-1)-7 \end{array}\right. \newlineFind the 3rd  3^{\text {rd }} term in the sequence.

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Full solution

Q. {b(1)=−2b(n)=b(n−1)−7 \left\{\begin{array}{l} b(1)=-2 \\ b(n)=b(n-1)-7 \end{array}\right. \newlineFind the 3rd  3^{\text {rd }} term in the sequence.
  1. Identify First Term: Identify the first term in the sequence.\newlineThe first term is given as b(1)=−2b(1) = -2.
  2. Find Second Term: Use the recursive formula to find the second term.\newlineThe recursive formula is b(n)=b(n−1)−7b(n) = b(n-1) - 7. To find the second term, we substitute nn with 22.\newlineb(2)=b(2−1)−7b(2) = b(2-1) - 7\newlineb(2)=b(1)−7b(2) = b(1) - 7\newlineb(2)=−2−7b(2) = -2 - 7\newlineb(2)=−9b(2) = -9
  3. Find Third Term: Use the recursive formula to find the third term.\newlineNow we use the second term to find the third term.\newlineb(3)=b(3−1)−7b(3) = b(3-1) - 7\newlineb(3)=b(2)−7b(3) = b(2) - 7\newlineb(3)=−9−7b(3) = -9 - 7\newlineb(3)=−16b(3) = -16

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