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{[d(1)=3],[d(n)=d(n-1)-14]:} Find the 3^("rd ") term in the sequence.

{d(1)=3d(n)=d(n−1)−14 \left\{\begin{array}{l} d(1)=3 \\ d(n)=d(n-1)-14 \end{array}\right. \newlineFind the 3rd  3^{\text {rd }} term in the sequence.

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Full solution

Q. {d(1)=3d(n)=d(n−1)−14 \left\{\begin{array}{l} d(1)=3 \\ d(n)=d(n-1)-14 \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 d(1)=3d(1) = 3.
  2. Determine Common Difference: Determine the common difference in the sequence.\newlineThe common difference is given by the recursive formula d(n)=d(n−1)−14d(n) = d(n-1) - 14, which means we subtract 1414 from the previous term to get the next term.
  3. Calculate Second Term: Calculate the second term using the common difference.\newlineThe second term is d(2)=d(1)−14=3−14=−11d(2) = d(1) - 14 = 3 - 14 = -11.
  4. Calculate Third Term: Calculate the third term using the common difference.\newlineThe third term is d(3)=d(2)−14=−11−14=−25d(3) = d(2) - 14 = -11 - 14 = -25.

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