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{[a(1)=20],[a(n)=a(n-1)*(3)/(2)]:}
What is the 
3^("rd ") term in the sequence?

{a(1)=20a(n)=a(n1)32 \left\{\begin{array}{l} a(1)=20 \\ a(n)=a(n-1) \cdot \frac{3}{2} \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?

Full solution

Q. {a(1)=20a(n)=a(n1)32 \left\{\begin{array}{l} a(1)=20 \\ a(n)=a(n-1) \cdot \frac{3}{2} \end{array}\right. \newlineWhat is the 3rd  3^{\text {rd }} term in the sequence?
  1. Define Recursive Sequence: The sequence is defined recursively, with the first term a(1)a(1) given as 2020. The nnth term a(n)a(n) is defined in terms of the previous term a(n1)a(n-1) as a(n)=a(n1)×(32)a(n) = a(n-1) \times (\frac{3}{2}). To find the third term a(3)a(3), we need to find the second term a(2)a(2) first, using the first term a(1)a(1). a(2)=a(1)×(32)=20×(32)=30a(2) = a(1) \times (\frac{3}{2}) = 20 \times (\frac{3}{2}) = 30.
  2. Calculate Second Term: Now that we have the second term a(2)a(2), we can find the third term a(3)a(3) by applying the recursive formula again.a(3)=a(2)×(32)=30×(32)=45.a(3) = a(2) \times \left(\frac{3}{2}\right) = 30 \times \left(\frac{3}{2}\right) = 45.

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