1. Chapter 9 Class 11 Sequences and Series
2. Serial order wise

Transcript

Misc 18 If a and b are the roots of x2 โ 3x + p = 0 and c,d are roots of x2 โ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = ๐/๐ & sum of roots = (โ๐)/๐ Misc 18 If a and b are the roots of x2 โ 3x + p = 0 and c,d are roots of x2 โ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = ๐/๐ & sum of roots = (โ๐)/๐ Misc 18 If a and b are the roots of x2 โ 3x + p = 0 and c,d are roots of x2 โ 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q โ p) = 17:15. We know that a, ar , ar2 , ar3, โฆ. 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 (๐ + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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 (๐ + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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 (๐ + ๐)/(๐ โ ๐) = 17/15 Taking L.H.S (๐ + ๐)/(๐ โ ๐) Putting value of p = ab & q = cd from (2) & (4) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), we need to find r first. = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), we need to find r first. Now Dividing (1) & (3) (๐ + ๐)/(๐ + ๐) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (๐ + ๐๐)/(๐๐2 +๐๐3) = 3/12 (๐(1 + ๐))/(๐๐2(1 + ๐)) = 3/12 1/๐2 = 3/12 1/๐2 = 1/4 r2 = 4 Now, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 โ 1) = (16 + 1)/(16 โ 1) = (17 )/15 = R.H.S Thus, L.H.S = R.H.S Hence proved