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QJEEM25Ma8fda: JEE Main 2025 Mathematics - Class 11 JEEM25Ma8fda Solution: The roots of the quadratic equation are and terms of an…

The roots of the quadratic equation 3x2−px+q=03 x^2-p x+q=0 are 10th 10^{\text {th }} and 11th 11^{\text {th }} terms of an arithmetic progression with common difference 32\frac{3}{2} . If the sum of the first 11 terms of this arithmetic progression is 88 , then q−2pq-2 p is equal to ________ .

Step-by-Step Verified Solution

amp;amp;S11=112(2a+10d)=88amp;a+5d=8amp;a=8−5×32=12amp; Roots are amp;amp;T10=a+9 d=12+9×32=14amp; T11=a+10 d=12+10×32=312amp;p3=T10+T11=14+312=592amp;p=1772amp;q3=T10×T11=7×31=217amp;q=651amp;q−2pamp;=651−177amp;=474\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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