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`        Plz answer this question I tried this but not getting the solution `
one year 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)^2however, 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 :)
```
one year ago
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