0

Guest

Courses New

Plz answer this question I tried this but not getting the solution

Plz answer this question I tried this but not getting the solution 

Question Image
Grade:11

1 Answers

Aditya Gupta
2081 Points
4 years ago
hello priyanka. the above ques is a direct application of cauchy schwartz inequality for real numbers (you can google this if you dont know about this inequality).
from cauchy schwartz inequality, we can say that
(x1^2 + x2^2 + …. + x(n-1)^2)*(x2^2 + x3^2 + …. + xn^2) is greater than or equal to (x1x2 + x3x3 + …..... + x(n-1)xn)^2.
however, in the ques we are given the inequality:
(x1^2 + x2^2 + …. + x(n-1)^2)*(x2^2 + x3^2 + …. + xn^2) is less than or equal to (x1x2 + x3x3 + …..... + x(n-1)xn)^2.
so obviously for both ineqs to be true, the given expressions are bound to be equal to each other. hence
(x1^2 + x2^2 + …. + x(n-1)^2)*(x2^2 + x3^2 + …. + xn^2) = (x1x2 + x3x3 + …..... + x(n-1)xn)^2
however, by cauchy ineq., equality holds if and only if ai= kbi for some constant k.
note that here ai= xi and bi= x(i+1)
so that xi= kx(i+1)
or x(i+1)/xi= 1/k = constant (say r).
but this is the very definition of a GP.
hence, x1, x2, …......., xn form a geometric progression.
kindly approve :)
 
 

Think You Can Provide A Better Answer ?

Other Related Questions on Analytical Geometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More