# Misc 18

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 18 If a and b are the roots of x2 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = 𝑐/𝑎 & sum of roots = (−𝑏)/𝑎 Misc 18 If a and b are the roots of x2 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. Introduction For quadratic equation ax2 + bx + c = 0 Product of roots = 𝑐/𝑎 & sum of roots = (−𝑏)/𝑎 Misc 18 If a and b are the roots of x2 – 3x + p = 0 and c,d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15. We know that a, ar , ar2 , ar3, …. 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 (𝑞 + 𝑝)/(𝑞 − 𝑝) = 17/15 Taking L.H.S (𝑞 + 𝑝)/(𝑞 − 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + 𝑎𝑏)/(𝑐𝑑 − 𝑎𝑏) We know that a, ar , ar2 , ar3, …. 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 (𝑞 + 𝑝)/(𝑞 − 𝑝) = 17/15 Taking L.H.S (𝑞 + 𝑝)/(𝑞 − 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + 𝑎𝑏)/(𝑐𝑑 − 𝑎𝑏) We know that a, ar , ar2 , ar3, …. 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 (𝑞 + 𝑝)/(𝑞 − 𝑝) = 17/15 Taking L.H.S (𝑞 + 𝑝)/(𝑞 − 𝑝) Putting value of p = ab & q = cd from (2) & (4) = (𝑐𝑑 + 𝑎𝑏)/(𝑐𝑑 − 𝑎𝑏) = (𝑐𝑑 + 𝑎𝑏)/(𝑐𝑑 − 𝑎𝑏) Putting values b = ar , c = ar2 , d = ar3 = ((𝑎𝑟^2 )(𝑎𝑟^3 ) + 𝑎(𝑎𝑟))/((𝑎𝑟^2 )(𝑎𝑟^3 ) − 𝑎(𝑎𝑟)) = (𝑎2𝑟4 + 𝑎2𝑟)/(𝑎2𝑟4 − 𝑎2𝑟) = (𝑎2𝑟4 + 𝑎2𝑟)/(𝑎2𝑟4 − 𝑎2𝑟) = (𝑎2𝑟(𝑟4 + 1))/(𝑎2𝑟(𝑟4 − 1 )) = (𝑟4 + 1 )/(𝑟4 − 1) So, (𝑞 + 𝑝)/(𝑞 − 𝑝) = (𝑟4 + 1 )/(𝑟4 − 1), we need to find r first. = (𝑐𝑑 + 𝑎𝑏)/(𝑐𝑑 − 𝑎𝑏) Putting values b = ar , c = ar2 , d = ar3 = ((𝑎𝑟^2 )(𝑎𝑟^3 ) + 𝑎(𝑎𝑟))/((𝑎𝑟^2 )(𝑎𝑟^3 ) − 𝑎(𝑎𝑟)) = (𝑎2𝑟4 + 𝑎2𝑟)/(𝑎2𝑟4 − 𝑎2𝑟) = (𝑎2𝑟4 + 𝑎2𝑟)/(𝑎2𝑟4 − 𝑎2𝑟) = (𝑎2𝑟(𝑟4 + 1))/(𝑎2𝑟(𝑟4 − 1 )) = (𝑟4 + 1 )/(𝑟4 − 1) So, (𝑞 + 𝑝)/(𝑞 − 𝑝) = (𝑟4 + 1 )/(𝑟4 − 1), we need to find r first. Now Dividing (1) & (3) (𝑎 + 𝑏)/(𝑐 + 𝑑) = 3/12 Putting values b = ar , c = ar2 , d = ar3 (𝑎 + 𝑎𝑟)/(𝑎𝑟2 +𝑎𝑟3) = 3/12 (𝑎(1 + 𝑟))/(𝑎𝑟2(1 + 𝑟)) = 3/12 1/𝑟2 = 3/12 1/𝑟2 = 1/4 r2 = 4 Now, (𝑞 + 𝑝)/(𝑞 − 𝑝) = (𝑟4 + 1 )/(𝑟4 − 1), Putting r2 = 4 = (4^2 + 1 )/(4^2 − 1) = (16 + 1)/(16 − 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

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .