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Question

Find the nth term of sequence  -3, 0, 5, ...
  1. 1 - 4n
  2. n - 2
  3. n2 - 4
  4. n2 + 1


Solution

The correct option is C n2 - 4
T1 = 12 - 4 = 1 - 4 = -3    OR
     = (1-2)×(1+2) = (-1)×3 = -3
T2 = 22 - 4 = 4 - 4 = 0    OR
     = (2-2)×(2+2) = 0×4 = 0
T3 = 32 - 4 = 9 - 4 = 5    OR
     = (3-2)×(3+2) = 1×5 = 5
thus 
Tn = n2 - 4           OR
     = (n-2)×(n+2)
     = n2 + 2n - 2n - 4 = n2 - 4

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