The roots of the quadratic equation 3 x ^2- p | vedclass

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Question

$S_{11}=\frac{11}{2}(2 a+10 d)=88$
$a+5 d=8$
$a=8-5 	imes \frac{3}{2}=\frac{1}{2}$
Roots are
$T_{10}=a+9 d=\frac{1}{2}+9 	imes \frac{3}{2}=14$

Vedclass · Accepted Answer

$474$

Vedclass · Answer

$S_{11}=\frac{11}{2}(2 a+10 d)=88$
$a+5 d=8$
$a=8-5 	imes \frac{3}{2}=\frac{1}{2}$
Roots are
$T_{10}=a+9 d=\frac{1}{2}+9 	imes \frac{3}{2}=14$
$T_{11}=a+10 d=\frac{1}{2}+10 	imes \frac{3}{2}=\frac{31}{2}$
$\frac{p}{3}=T_{10}+T_{11}=14+\frac{31}{2}=\frac{59}{2}$
$p=\frac{177}{2}$
$\frac{q}{3}=T_{10} 	imes T_{11}=7 	imes 31=217$
$q=651$
$q-2 p$
$=651-177$
$=474$

The roots of the quadratic equation $3 x ^2- px + q =0$ 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_______

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