Question ID: #347
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:
Solution:
Let the AP have first term $a$ and common difference $d = \frac{3}{2}$.
Sum of first 11 terms ($S_{11}$) is 88:
$$S_{11} = \frac{11}{2}[2a + (11-1)d] = 88$$
$$\frac{11}{2}[2a + 10(\frac{3}{2})] = 88$$
$$\frac{1}{2}[2a + 15] = 8$$
$$2a + 15 = 16 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}$$
The roots of the quadratic equation are $T_{10}$ and $T_{11}$.
$$T_{10} = a + 9d = \frac{1}{2} + 9(\frac{3}{2}) = \frac{1}{2} + \frac{27}{2} = \frac{28}{2} = 14$$
$$T_{11} = a + 10d = \frac{1}{2} + 10(\frac{3}{2}) = \frac{1}{2} + 15 = \frac{31}{2}$$
The quadratic equation with roots 14 and $\frac{31}{2}$ is:
$$(x – 14)(x – \frac{31}{2}) = 0$$
$$x^2 – (14 + \frac{31}{2})x + (14 \cdot \frac{31}{2}) = 0$$
$$x^2 – \frac{59}{2}x + 217 = 0$$
Multiply by 3 to match the form $3x^2 – px + q = 0$:
$$3x^2 – 3(\frac{59}{2})x + 3(217) = 0$$
$$3x^2 – \frac{177}{2}x + 651 = 0$$
(Note: The problem implies integer coefficients usually, but let’s compare directly to $3x^2 – px + q = 0$).
$$p = \frac{177}{2}, \quad q = 651$$
We need to find $q – 2p$:
$$q – 2p = 651 – 2(\frac{177}{2})$$
$$q – 2p = 651 – 177$$
$$q – 2p = 474$$
Ans. (474)
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