Question

Help me with this, please....
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Answer 1

$\begin{gathered}\begin{gathered}\bf Given : - \begin{cases} &\sf{ \alpha \: and \: \beta \: are \: roots \: of \: {x}^{2} - 2x + A = 0} \\ &\sf{ \gamma \: and \: \delta \: are \: the \: roots \: of \: {x}^{2} - 18x + B = 0}\\ &\sf{ \alpha, \beta , \gamma ,\delta, \: are \: in \: AP } \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To Find - \begin{cases} &\sf{valuesof \: A \: and \: B} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \: \underline \red{Step :- 1}$

$\bf \: \red{According \: to \: statement }$

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$\sf \:  ⟼ \alpha , \beta , \gamma ,\delta \: are \: in \: AP$

$\bf\implies \: \alpha = a - 3d$

$\bf\implies \: \beta = a - d$

$\bf\implies \: \gamma = a + d$

$\bf\implies \:\delta = a + 3d$

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$\bf \: \underline \red{Step :- 2}$

$\sf \:  ⟼\alpha \: and \: \beta \: are \: roots \: of \: {x}^{2} - 2x + A = 0$

$\bf\implies \: \alpha + \beta = 2 \:\sf \:  ⟼(1)$

$\bf\implies \: \alpha \beta = A\sf \:  ⟼(2)$

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$\bf \: \underline \red{Step :- 3}$

$\sf \:  ⟼ \gamma \: and \: \delta \: are \: the \: roots \: of \: {x}^{2} - 18x + B = 0$

$\bf\implies \: \gamma + \delta = 18\sf \:  ⟼(3)$

$\bf\implies \: \gamma \delta = B\sf \:  ⟼(4)$

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$\bf \: \underline \red{Step :- 4}$

On adding equation (1) and equation (3), we get

$\sf \:  ⟼ \alpha + \beta + \gamma + \delta = 2 + 18$

$\sf \:  ⟼a - 3d + a - d+ a+ d + a + 3d = 20$

$\sf \:  ⟼4a = 20$

$\bf\implies \:a = 5 \: \sf \:  ⟼(5)$

From equation (1), we have

$\sf \:  ⟼ \alpha + \beta = 2$

$\sf \:  ⟼a - 3d +a - d= 2$

$\sf \:  ⟼2a - 4d = 2$

$\sf \:  ⟼a - 2d = 1$

On substituting the value of a = 5, from (1), we get

$\sf \:  ⟼5 - 2d = 1$

$\sf \:  ⟼ - 2d = - 4$

$\bf\implies \:d = 2\sf \:  ⟼(6)$

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$\bf \: \underline \red{Step :- 5}$

From equation (2), we have

$\sf \:  ⟼ \alpha \beta = A$

$\sf \:  ⟼A = (a - 3d)(a - d)$

On substituting the values of a and d we, get

$\sf \:  ⟼A = (5 - 6)(5 - 2) = - 3$

$\bf\implies \:A = - 3\sf \:  ⟼(7)$

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$\bf \: \underline \red{Step :- 6}$

From equation (4), we have

$\sf \:  ⟼ \gamma \delta = B$

$\sf \:  ⟼B = (a + d)(a + 3d)$

On substituting the values of a and d we, get

$\sf \:  ⟼B =(5 + 2) (5 + 6) = 77$

$\bf\implies \:B = 77$

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$\large{\boxed{\boxed{\bf{Hence, A = - 3 \: and \: B = 77}}}}$

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