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

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Question

(A) Given the sum of the first $11$ terms $S_{11} = 88$ and common difference $d = \frac{3}{2}$.Using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$:$88

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$474$

Vedclass · Answer

(A) Given the sum of the first $11$ terms $S_{11} = 88$ and common difference $d = \frac{3}{2}$.Using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$:$88 = \frac{11}{2}(2a + 10 	imes \frac{3}{2})$$8 = \frac{1}{2}(2a + 15) \implies 16 = 2a + 15 \implies 2a = 1 \implies a = \frac{1}{2}$.The $10^{	ext{th}}$ and $11^{	ext{th}}$ terms are:$T_{10} = a + 9d = \frac{1}{2} + 9(\frac{3}{2}) = \frac{1}{2} + \frac{27}{2} = 14$.$T_{11} = a + 10d = \frac{1}{2} + 10(\frac{3}{2}) = \frac{1}{2} + 15 = \frac{31}{2}$.For the quadratic equation $3x^2 - px + q = 0$,the sum of roots is $\frac{p}{3} = T_{10} + T_{11} = 14 + \frac{31}{2} = \frac{59}{2} \implies p = \frac{177}{2}$.The product of roots is $\frac{q}{3} = T_{10} 	imes T_{11} = 14 	imes \frac{31}{2} = 7 	imes 31 = 217 \implies q = 651$.Calculating $q - 2p = 651 - 2(\frac{177}{2}) = 651 - 177 = 474$.

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 a common difference $d = \frac{3}{2}$. If the sum of the first $11$ terms of this arithmetic progression is $88$,then $q - 2p$ is equal to:

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