Last updated at May 29, 2018 by Teachoo

Transcript

Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. Introduction If x2 + y2 + z2 0 So, if x2 + y2 + z2 0, x2 + y2 + z2 = 0 i.e. x = 0, y = 0 , z = 0 Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. We need to show a, b, c & d are in GP So, common ratio must be same we have to prove / = / = / It is given that (a2 + b2 + c2) p2 2 (ab + bc + cd) p + (b2 + c2 + d2) 0 Solving a2p2 + b2p2 + c2p2 2abp 2bcp 2cdp + b2 + c2 + d2 0 (ap)2 + b2 2abp + (bp)2 + c2 2bcp + (cp)2 + d2 2cdp 0 Example 23, If a, b, c, d and p are different real numbers such that (a2 + b2 + c2)p2 2(ab + bc + cd) p + (b2 + c2 + d2) 0, then show that a, b, c and d are in G.P. We need to show a, b, c & d are in GP So, common ratio must be same we have to prove / = / = / It is given that (a2 + b2 + c2) p2 2 (ab + bc + cd) p + (b2 + c2 + d2) 0 Solving a2p2 + b2p2 + c2p2 2abp 2bcp 2cdp + b2 + c2 + d2 0 (ap)2 + b2 2abp + (bp)2 + c2 2bcp + (cp)2 + d2 2cdp 0 Solving (ap b) = 0 ap = b / = p Solving (bp c) = 0 bp = c / = p Solving (cp d) = 0 cp = d / = p This implies that (b )/a = (c )/b = (d )/c = p Hence a, b, c and d are in G.P.

Chapter 9 Class 11 Sequences and Series

Example 6
Important

Ex 9.2, 5 Important

Ex 9.2, 9 Important

Ex 9.2, 15 Important

Ex 9.2, 17 Important

Example 14 Important

Example 15 Important

Ex 9.3, 3 Important

Ex 9.3, 11 Important

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 22 Important

Ex 9.3, 28 Important

Ex 9.3, 29 Important

Ex 9.4.4 Important

Ex 9.4, 7 Important

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Example 23 Important You are here

Misc 16 Important

Misc 19 Important

Misc 25 Important

Misc 28 Important

Misc 32 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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.