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Let alpha1,beta1 be the roots x^2-6x+p=0...

Let `alpha_1,beta_1` be the roots `x^2-6x+p=0a n d` `alpha_2,beta_2` be the roots `x^2-54 x+q=0dot` If `alpha_1,beta_1,alpha_2,beta_2` form an increasing G.P., then sum of the digits of the value of `(q-p)` is ___________.

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Verified by Experts

The correct Answer is:
540

Let `alpha_(1)=A,beta_(1)=AR,alpha_(2)=AR^(2),beta_(2)=AR^(3)`
we have `alpha_(1)+beta_(2)=6impliesA(1+R)=6" "(1)`
`alpha_(1)beta_(1)=pimpliesA^(2)R=p" "(2)`
Also `alpha_(2)+beta_(2)=54impliesAR^(2)(1+R)=54" "(3)`
`alpha_(2)beta_(2)=qimpliesA^(2)R^(5)=q" "(4)`
Now, on dividing Eq. (3) by Eq. (1), we get
`(AR^(2)(1+R))/(A(1+R))=(54)/(6)=9orR^(2)=9`
`:.`R=3 (as it is an increasing G.P.)
`:.` On putting R=3 in Eq. (1), we get
`A=(6)/(4)=(3)/(2)`
`:.p=A^(2)R=(9)/(4)xx3=(27)/(4)`
and `q=A^(3)R^(5)=(9)/(4)xx243=(2187)/(4)`
Hence, `q-p=(2187-27)/(4)=(2160)/(4)=540`
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