The roots of the quadratic equation $$3x^{2} - px + q = 0$$ are $$10^{th}$$ and $$11^{th}$$ 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 equal to

Let the roots of $$3x^{2} - px + q = 0$$ be $$\alpha$$ and $$\beta$$. 

It is given that $$\alpha$$ and $$\beta$$ are the 10th and 11th term of an A.P with a common difference of $$\dfrac{3}{2}$$.

10th term = $$a+9d$$ = $$a+\left(9\times\dfrac{3}{2}\right)$$ = $$a+\dfrac{27}{2}$$

11th term = $$a+10d$$ = $$a+\left(10\times\dfrac{3}{2}\right)$$ = $$a+15$$

Now, it is given that the sum of the first 11 terms of the A.P is 88. 

$$\dfrac{11}{2}\times\left(a+\left(a+15\right)\right)=88$$

$$2a+15=16$$

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

10th term = $$\dfrac{1}{2}+\dfrac{27}{2}=14$$

11th term = $$\dfrac{1}{2}+15=\dfrac{31}{2}$$

So, $$\alpha=14$$ and $$\beta=\dfrac{31}{2}$$

Now, in $$3x^{2} - px + q = 0$$, 

$$\alpha+\beta=\dfrac{p}{3}$$

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

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

$$p=\dfrac{59\times3}{2}$$

$$\alpha\times\beta=\dfrac{q}{3}$$

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

$$q=7\times31\times3$$

$$q=651$$

Now, $$q-2p$$ = $$651-\left(2\times\dfrac{59\times3}{2}\right)$$

$$q-2p$$ = $$651-177=474$$

Hence, the value of $$(q-2p)$$ is 474. 

$$\therefore\ $$ The required answer is 474.