Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
The roots of the quadratic equation $3x^{2}-px+q=0$ are the $10^{\text{th}}$ and $11^{\text{th}}$ terms of an arithmetic progression with common difference $\dfrac{3}{2}$. If the sum of the first $11$ terms of this arithmetic progression is $88$, then $q-2p$ is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $S_{11}=\dfrac{11}{2}(2a+10d)\Rightarrow a+5d=8$ pins down $a$. Then $T_{10},T_{11}$ give the sum and product of the quadratic's roots via Vieta's $p/3$ and $q/3$.
$S_{11}=\dfrac{11}{2}(2a+10d)=88\Longrightarrow a+5d=8.$ With $d=\tfrac{3}{2}$: $a=8-\tfrac{15}{2}=\tfrac{1}{2}$.
$$T_{10}=a+9d=\tfrac{1}{2}+\tfrac{27}{2}=14,\qquad T_{11}=a+10d=\tfrac{1}{2}+15=\tfrac{31}{2}.$$
By Vieta's, sum $=\dfrac{p}{3}=T_{10}+T_{11}=14+\tfrac{31}{2}=\tfrac{59}{2}\Rightarrow p=\tfrac{177}{2}$.
Product $=\dfrac{q}{3}=T_{10}T_{11}=14\cdot\tfrac{31}{2}=217\Rightarrow q=651.$
$$q-2p=651-177=474.$$
Correct Answer: 474