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 $\frac{3}{2}$. If the sum of the first $11$ terms of this arithmetic progression is $88$, then $q-2p$ is:

Let the first term of the arithmetic progression be $a$.

Common difference is:

$d=\frac{3}{2}$

Sum of first $11$ terms is:

$S_{11}=88$

Formula for sum of first $n$ terms:

$S_n=\frac{n}{2}[2a+(n-1)d]$

So,

$88=\frac{11}{2}[2a+10d]$

Substitute $d=\frac{3}{2}$:

$88=\frac{11}{2}\left[2a+10\times \frac{3}{2}\right]$

$88=\frac{11}{2}[2a+15]$

Now,

$2a+15=16$

$2a=1$

$a=\frac{1}{2}$

The $10^{\text{th}}$ term is:

$T_{10}=a+9d$

$T_{10}=\frac{1}{2}+9\times \frac{3}{2}$

$T_{10}=\frac{1}{2}+\frac{27}{2}$

$T_{10}=14$

The $11^{\text{th}}$ term is:

$T_{11}=a+10d$

$T_{11}=\frac{1}{2}+10\times \frac{3}{2}$

$T_{11}=\frac{1}{2}+15$

$T_{11}=\frac{31}{2}$

So, the roots of $3x^2-px+q=0$ are $14$ and $\frac{31}{2}$.

For equation $3x^2-px+q=0$,

Sum of roots:

$\frac{p}{3}=14+\frac{31}{2}$

$\frac{p}{3}=\frac{28+31}{2}$

$\frac{p}{3}=\frac{59}{2}$

$p=\frac{177}{2}$

Product of roots:

$\frac{q}{3}=14\times \frac{31}{2}$

$\frac{q}{3}=217$

$q=651$

Now,

$q-2p=651-2\times \frac{177}{2}$

$q-2p=651-177$

$q-2p=474$

Therefore, the correct answer is option $2$.Given,