Last updated at May 29, 2018 by Teachoo

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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 = (โ๐)/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 = (โ๐)/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, โฆ. 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) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) We know that a, ar , ar2 , ar3, โฆ. 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) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 โ 1), we need to find r first. = (๐๐ + ๐๐)/(๐๐ โ ๐๐) Putting values b = ar , c = ar2 , d = ar3 = ((๐๐^2 )(๐๐^3 ) + ๐(๐๐))/((๐๐^2 )(๐๐^3 ) โ ๐(๐๐)) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐4 + ๐2๐)/(๐2๐4 โ ๐2๐) = (๐2๐(๐4 + 1))/(๐2๐(๐4 โ 1 )) = (๐4 + 1 )/(๐4 โ 1) So, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 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 + ๐๐)/(๐๐2 +๐๐3) = 3/12 (๐(1 + ๐))/(๐๐2(1 + ๐)) = 3/12 1/๐2 = 3/12 1/๐2 = 1/4 r2 = 4 Now, (๐ + ๐)/(๐ โ ๐) = (๐4 + 1 )/(๐4 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

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18 You are here

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.