Question

9. Given that ⁿÄ₄ (nC4), ⁿÄ₅ (nC5) and ⁿÄ₆ (nC6) are the first 3 terms of a linear sequence (A.P.), find the:
(a) values of n;
(b) common difference of the sequence.

Accepted Answer

For an A.P.: 2(nC5) = nC4+nC6. Using the ratios nC5/nC4=(n−4)/5 and nC6/nC5=(n−5)/6 and simplifying leads to the quadratic n²−21n+98=0, giving n=(21±7)/2 = 14 or 7.
(a) n = 7 or n = 14
(b) For n=7: nC4=35, nC5=21, nC6=7, common difference d=−14.
For n=14: nC4=1001, nC5=2002, nC6=3003, common difference d=1001.

WAEC Further Mathematics 2022 Theory — Question 9

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