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{[a(1)=(5)/(3)],[a(n)=a(n-1)*(-9)]:}
What is the 
3^("rd ") term in the sequence?

{a(1)=53a(n)=a(n1)(9) \left\{\begin{array}{l} a(1)=\frac{5}{3} \\ a(n)=a(n-1) \cdot(-9) \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?

Full solution

Q. {a(1)=53a(n)=a(n1)(9) \left\{\begin{array}{l} a(1)=\frac{5}{3} \\ a(n)=a(n-1) \cdot(-9) \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?
  1. Identify first term and common ratio: Identify the first term of the sequence and the common ratio.\newlineThe first term a(1)a(1) is given as 53\frac{5}{3}. The sequence is defined recursively, with each term being 9-9 times the previous term. Therefore, the common ratio is 9-9.
  2. Calculate second term: Calculate the second term using the first term and the common ratio.\newlineThe second term a(2)a(2) is a(1)a(1) multiplied by the common ratio.\newlinea(2)=a(1)×(9)a(2) = a(1) \times (-9)\newlinea(2)=(53)×(9)a(2) = \left(\frac{5}{3}\right) \times (-9)\newlinea(2)=15a(2) = -15
  3. Calculate third term: Calculate the third term using the second term and the common ratio.\newlineThe third term a(3)a(3) is a(2)a(2) multiplied by the common ratio.\newlinea(3)=a(2)×(9)a(3) = a(2) \times (-9)\newlinea(3)=15×(9)a(3) = -15 \times (-9)\newlinea(3)=135a(3) = 135

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