JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 24)
The roots of the quadratic equation $3 x^2-p x+q=0$ are $10^{\text {th }}$ and $11^{\text {th }}$ terms of an arithmetic progression with common difference $\frac{3}{2}$. If the sum of the first 11 terms of this arithmetic progression is 88 , then $q-2 p$ is equal to ________ .
Answer
474
Explanation
$$\begin{aligned}
&\begin{aligned}
& S_{11}=\frac{11}{2}(2 a+10 d)=88 \\
& a+5 d=8 \\
& a=8-5 \times \frac{3}{2}=\frac{1}{2}
\end{aligned}\\
&\text { Roots are }\\
&\begin{aligned}
& \mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=\frac{1}{2}+9 \times \frac{3}{2}=14 \\
& \mathrm{~T}_{11}=\mathrm{a}+10 \mathrm{~d}=\frac{1}{2}+10 \times \frac{3}{2}=\frac{31}{2} \\
& \frac{\mathrm{p}}{3}=\mathrm{T}_{10}+\mathrm{T}_{11}=14+\frac{31}{2}=\frac{59}{2} \\
& \mathrm{p}=\frac{177}{2} \\
& \frac{\mathrm{q}}{3}=\mathrm{T}_{10} \times \mathrm{T}_{11}=7 \times 31=217 \\
& \mathrm{q}=651 \\
& \mathrm{q}-2 \mathrm{p} \\
& =651-177 \\
& =474
\end{aligned}
\end{aligned}$$
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