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If 
a_(1)=2 and 
a_(n+1)=(a_(n))^(2)-2 then find the value of 
a_(3).
Answer:

If a1=2 a_{1}=2 and an+1=(an)22 a_{n+1}=\left(a_{n}\right)^{2}-2 then find the value of a3 a_{3} .\newlineAnswer:

Full solution

Q. If a1=2 a_{1}=2 and an+1=(an)22 a_{n+1}=\left(a_{n}\right)^{2}-2 then find the value of a3 a_{3} .\newlineAnswer:
  1. Given terms and formula: We are given the first term of the sequence, a1=2a_{1} = 2, and the recursive formula for the sequence, an+1=(an)22a_{n+1} = (a_{n})^2 - 2. To find a3a_{3}, we first need to find a2a_{2} using the recursive formula.
  2. Find a2a_{2}: Using the recursive formula an+1=(an)22a_{n+1} = (a_{n})^2 - 2, we substitute n=1n = 1 to find a2a_{2}:
    a2=(a1)22a_{2} = (a_{1})^2 - 2
    a2=(2)22a_{2} = (2)^2 - 2
    a2=42a_{2} = 4 - 2
    a2=2a_{2} = 2
  3. Find a3a_{3}: Now that we have a2a_{2}, we can use it to find a3a_{3} using the same recursive formula:\newlinea3=(a2)22a_{3} = (a_{2})^2 - 2\newlinea3=(2)22a_{3} = (2)^2 - 2\newlinea3=42a_{3} = 4 - 2\newlinea3=2a_{3} = 2

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