Further Mathematics Paper 2, Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 2

If the sum of the first n terms of the series 4 + 7 + 10 + ... is 209, find n. .

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Observation

This question was also reported to have been attempted by majority of the candidate and they performed well in it. However, some candidates were reported not to have applied the quadratic formula correctly when solving for n. They wrote n = - 5± √ 5²-12(-418)/2x3 instead of n =   -5√±5²-12(-418)/2x3 . Hence the obtained wrong values for n.
Candidates were expected to respond as follows:
S = 209, a = 4, d = 3. Therefore, substituting these into the formula, Sn = n/2[2a + (n -1)d], we obtained Sn = n/2(8 + (n -1 )3) = 209. => 3n² + 5n - 418 = O. By factorizing this equation we obtained (3n + 38)(n -11) = 0 => n = 11. The other solution - 38was not accepted because n cannot be a fraction or a negative number.
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