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JAMB 2012 · UTME · Q19

The nnth term of a sequence is n2−6n−4n^2 - 6n - 4. Find the sum of the 3rd and 4th terms.

Worked solution (try it first)
  1. Put n=3n = 3: T3=9−18−4=−13T_3 = 9 - 18 - 4 = -13.
  2. Put n=4n = 4: T4=16−24−4=−12T_4 = 16 - 24 - 4 = -12.
  3. So the sum is −13+(−12)=−25-13 + (-12) = -25, option D.

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