Chapter 9 Class 11 Sequences and Series
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams You are here
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 17 Deleted for CBSE Board 2024 Exams
Example 9 Important
Example 10 Important
Ex 8.2, 3 Important
Ex 8.2, 11 Important
Ex 8.2, 17 Important
Ex 8.2, 18 Important
Ex 8.2, 22 Important
Ex 8.2, 28
Ex 8.2, 29 Important
Ex 9.4.4 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams You are here
Question 10 Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams You are here
Question 9 Important Deleted for CBSE Board 2024 Exams You are here
Misc 10 Important
Question 13 Important Deleted for CBSE Board 2024 Exams
Misc 14 Important
Misc 18 Important
Chapter 9 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Question 9, 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 Question 9, 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 Question 9, 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.