The roots of the equation $3x^2-px+q=0$ are the $10^{\text{th}}$ and $11^{\text{th}}$ terms of an AP with common difference $\dfrac{3}{2}$ and $S_{11}=88$. Then $q-2p$ equals
Step-by-Step Solution
Key Concept: Use $S_{11}=11a_6$ to find $a_6$, compute $a_{10}$ and $a_{11}$, then apply Vieta's formulas.
$S_{11}=11a_6=88\Rightarrow a_6=8$. With $d=\frac{3}{2}$: $a_1=a_6-5d=8-\frac{15}{2}=\frac{1}{2}$.
$a_{10}=\frac{1}{2}+9\cdot\frac{3}{2}=\frac{1}{2}+\frac{27}{2}=14$.
$a_{11}=\frac{1}{2}+10\cdot\frac{3}{2}=\frac{31}{2}$.
By Vieta's: $\dfrac{p}{3}=14+\dfrac{31}{2}=\dfrac{59}{2}\Rightarrow p=\dfrac{177}{2}$.
$\dfrac{q}{3}=14\times\dfrac{31}{2}=217\Rightarrow q=651$.
$q-2p=651-177=474$.
Correct Answer: 474