The roots of the quadratic equation 3x2 – px + q = 0 are 10th and 11th terms of an arithmetic progression with common difference 32. If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2q is equal to _________.

S11=112(2a+10d)=88a+5d=8a=8−5×32=12Roots areT10=a+9d=12+9×32=14T11=a+10d=12+10×32=312p3=T10+T11=14+312=592p=1772q3=T10×T11=7×31=217q=651q−2p=651−177=474

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The roots of the quadratic equation 3x² – 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 – 2q is equal to _________.

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answer is 474.

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Detailed Solution

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