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{[a(1)=-11],[a(n)=a(n-1)*10]:}
What is the 
4^("th ") term in the sequence?

{a(1)=11a(n)=a(n1)10 \left\{\begin{array}{l} a(1)=-11 \\ a(n)=a(n-1) \cdot 10 \end{array}\right. \newlineWhat is the 4th  4^{\text {th }} term in the sequence?

Full solution

Q. {a(1)=11a(n)=a(n1)10 \left\{\begin{array}{l} a(1)=-11 \\ a(n)=a(n-1) \cdot 10 \end{array}\right. \newlineWhat is the 4th  4^{\text {th }} term in the sequence?
  1. Identify Sequence Type: Identify the type of sequence.\newlineThe given sequence is defined recursively, with each term being 1010 times the previous term. This indicates that the sequence is geometric.
  2. Find First Term: Determine the first term of the sequence.\newlineThe first term a(1)a(1) is given as 11-11.
  3. Find Second Term: Use the recursive formula to find the second term.\newlineThe second term a(2)a(2) is a(1)a(1) multiplied by 1010.\newlinea(2)=a(1)×10=11×10=110a(2) = a(1) \times 10 = -11 \times 10 = -110
  4. Find Third Term: Use the recursive formula to find the third term.\newlineThe third term a(3)a(3) is a(2)a(2) multiplied by 1010.\newlinea(3)=a(2)×10=(110)×10=1100a(3) = a(2) \times 10 = (-110) \times 10 = -1100
  5. Find Fourth Term: Use the recursive formula to find the fourth term.\newlineThe fourth term a(4)a(4) is a(3)a(3) multiplied by 1010.\newlinea(4)=a(3)×10=1100×10=11000a(4) = a(3) \times 10 = -1100 \times 10 = -11000

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