JAMB 2010 · UTME · Q22

The 3rd term of an arithmetic progression is −9-9 and the 7th term is −29-29. Find the 10th term of the progression.

Worked solution (try it first)
  1. From the 3rd term to the 7th term is 4 steps of dd: 4d=−29−(−9)=−204d = -29 - (-9) = -20, so d=−5d = -5.
  2. Go back two steps from the 3rd term: a=−9−2(−5)=1a = -9 - 2(-5) = 1.
  3. The 10th term is a+9d=1+9(−5)=−44a + 9d = 1 + 9(-5) = -44, option C.

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