Miscellaneous

Chapter 9 Class 11 Sequences and Series
Serial order wise

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Misc 18 If a and b are the roots of x2 β 3x + p = 0 and c,d are roots of x2 β 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q β p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = π/π & sum of roots = (βπ)/π Misc 18 If a and b are the roots of x2 β 3x + p = 0 and c,d are roots of x2 β 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q β p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = π/π & sum of roots = (βπ)/π Misc 18 If a and b are the roots of x2 β 3x + p = 0 and c,d are roots of x2 β 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q β p) = 17:15. We know that a, ar , ar2 , ar3, β¦. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (π + π)/(π β π) = 17/15 Taking L.H.S (π + π)/(π β π) Putting value of p = ab & q = cd from (2) & (4) = (ππ + ππ)/(ππ β ππ) We know that a, ar , ar2 , ar3, β¦. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (π + π)/(π β π) = 17/15 Taking L.H.S (π + π)/(π β π) Putting value of p = ab & q = cd from (2) & (4) = (ππ + ππ)/(ππ β ππ) We know that a, ar , ar2 , ar3, β¦. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We have to prove (π + π)/(π β π) = 17/15 Taking L.H.S (π + π)/(π β π) Putting value of p = ab & q = cd from (2) & (4) = (ππ + ππ)/(ππ β ππ) = (ππ + ππ)/(ππ β ππ) Putting values b = ar , c = ar2 , d = ar3 = ((ππ^2 )(ππ^3 ) + π(ππ))/((ππ^2 )(ππ^3 ) β π(ππ)) = (π2π4 + π2π)/(π2π4 β π2π) = (π2π4 + π2π)/(π2π4 β π2π) = (π2π(π4 + 1))/(π2π(π4 β 1 )) = (π4 + 1 )/(π4 β 1) So, (π + π)/(π β π) = (π4 + 1 )/(π4 β 1), we need to find r first. = (ππ + ππ)/(ππ β ππ) Putting values b = ar , c = ar2 , d = ar3 = ((ππ^2 )(ππ^3 ) + π(ππ))/((ππ^2 )(ππ^3 ) β π(ππ)) = (π2π4 + π2π)/(π2π4 β π2π) = (π2π4 + π2π)/(π2π4 β π2π) = (π2π(π4 + 1))/(π2π(π4 β 1 )) = (π4 + 1 )/(π4 β 1) So, (π + π)/(π β π) = (π4 + 1 )/(π4 β 1), we need to find r first. Now Dividing (1) & (3) (π + π)/(π + π) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (π + ππ)/(ππ2 +ππ3) = 3/12 (π(1 + π))/(ππ2(1 + π)) = 3/12 1/π2 = 3/12 1/π2 = 1/4 r2 = 4 Now, (π + π)/(π β π) = (π4 + 1 )/(π4 β 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 β 1) = (16 + 1)/(16 β 1) = (17 )/15 = R.H.S Thus, L.H.S = R.H.S Hence proved