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{[b(1)=-7],[b(n)=b(n-1)+12]:}
Find the 
4^("th ") term in the sequence.

{b(1)=7b(n)=b(n1)+12 \left\{\begin{array}{l} b(1)=-7 \\ b(n)=b(n-1)+12 \end{array}\right. \newlineFind the 4th  4^{\text {th }} term in the sequence.

Full solution

Q. {b(1)=7b(n)=b(n1)+12 \left\{\begin{array}{l} b(1)=-7 \\ b(n)=b(n-1)+12 \end{array}\right. \newlineFind the 4th  4^{\text {th }} term in the sequence.
  1. Understand the sequence definition: Understand the sequence definition.\newlineThe sequence is defined recursively, with the first term b(1)b(1) given as 7-7. Each subsequent term is found by adding 1212 to the previous term. This means that to find the 44th term, we need to apply the recursive formula three times starting from the first term.
  2. Find the second term: Find the second term using the recursive formula.\newlineWe know that b(1)=7b(1) = -7. To find b(2)b(2), we use the formula b(n)=b(n1)+12b(n) = b(n-1) + 12.\newlineb(2)=b(1)+12=7+12=5b(2) = b(1) + 12 = -7 + 12 = 5.
  3. Find the third term: Find the third term using the recursive formula.\newlineNow that we have b(2)b(2), we can find b(3)b(3) by adding 1212 to b(2)b(2).\newlineb(3)=b(2)+12=5+12=17b(3) = b(2) + 12 = 5 + 12 = 17.
  4. Find the fourth term: Find the fourth term using the recursive formula.\newlineFinally, we find b(4)b(4) by adding 1212 to b(3)b(3).\newlineb(4)=b(3)+12=17+12=29b(4) = b(3) + 12 = 17 + 12 = 29.

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