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Complete the recursive formula of the geometric sequence {:[-125","-25","-5","-1","dots.],[b(1)=],[b(n)=b(n-1)]:}

Complete the recursive formula of the geometric sequence\newline−125,−25,−5,−1,….b(1)=□b(n)=b(n−1)⋅□ \begin{array}{l} -125,-25,-5,-1, \ldots . \\ b(1)=\square \\ b(n)=b(n-1) \cdot \square \end{array}

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Full solution

Q. Complete the recursive formula of the geometric sequence\newline−125,−25,−5,−1,….b(1)=□b(n)=b(n−1)⋅□ \begin{array}{l} -125,-25,-5,-1, \ldots . \\ b(1)=\square \\ b(n)=b(n-1) \cdot \square \end{array}
  1. Identify First Term: Identify the first term of the sequence.\newlineThe first term of the sequence is given as b(1)=−125b(1) = -125.
  2. Determine Common Ratio: Determine the common ratio by dividing the second term by the first term.\newlineThe common ratio rr can be found by dividing the second term −25-25 by the first term −125-125.\newliner=−25−125=15r = \frac{-25}{-125} = \frac{1}{5}
  3. Write Recursive Formula: Write the recursive formula using the first term and the common ratio.\newlineThe recursive formula for a geometric sequence is b(n)=b(n−1)×rb(n) = b(n-1) \times r, where rr is the common ratio.\newlineSince we have b(1)=−125b(1) = -125 and r=15r = \frac{1}{5}, the recursive formula is:\newlineb(n)=b(n−1)×(15)b(n) = b(n-1) \times \left(\frac{1}{5}\right)

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