Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let <i>n</i> be the greatest integer for which <i>5p</i><sup>2</sup> <i>− 16, 2p, n − 2</i> are distinct consecutive terms of an AP, where <i>p</i> ∈ ℝ. If the common difference of the AP is \(\frac{m}{n}\), where <i>m, n</i> ∈ ℕ and <i>m, n</i> are relatively prime, the value of <i>m + n</i> is</p>
<p>(a) 133</p>
<p>(b) 138</p>
<p>(c) 143</p>
<p>(d) 148</p>
Step-by-Step Solution
Key Concept: Use the AP condition that the middle term is the average of the other two, then find the constraint on n using the discriminant condition for p to be real.
<p><strong>Step 1:</strong> Since $5p^2 - 16, 2p, n - 2$ are in AP, we have:</p><p>$$4p - (5p^2 - 16) = (n - 2) - 2p$$</p><p>$$4p - 5p^2 + 16 = n - 2 - 2p$$</p><p>$$2n - 4p - n^2 - 16 = 0$$</p><p><strong>Step 2:</strong> For $p$ to be real, the discriminant must be non-negative:</p><p>$$B^2 - 4AC \geq 0$$</p><p>$$16n^2 - 4(5)(n^2 - 16) \geq 0$$</p><p>$$-n^2 + 80 \geq 0$$</p><p>$$n^2 \leq 80$$</p><p>Therefore, $n \leq 8$ (greatest integer).</p><p><strong>Step 3:</strong> When $n = 8$, from $5p^2 - 32p + 48 = 0$:</p><p>$$(p - 4)(5p - 12) = 0$$</p><p>So $p = 4$ or $p = \frac{12}{5}$</p><p><strong>Step 4:</strong> The common difference is $d = n - 4p = \frac{138}{5}$, so $m = 138, n = 5$.</p><p>∴ <i>m + n</i> = 143</p>
Correct Answer: C