Let `x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5)` and `y_(1) lt y_(2) lt y_(3) lt y_(4) lt y_(5)` are in AP such that `underset(I = 1)overset(5)(sum) x_(i) = underset(I = 1)overset(5)(sum) y_(i) = 25` and `underset(I =1)overset(5)(II) x_(i) = underset(I = 1)overset(5)(II) y_(i) = 0` then `|y_(5) - x_(5)|`A. `beta_(n)" "and log" "beta_(n)`B. `K_(n)" "and log" "beta_(n-1)`C. `beta_(n-1)" "and log" "beta_(n-1)`D. `K_(n-1)" "and log" "K_(n-1)`

Let `x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5)` and `y_(1) lt y_(2) lt

1.	Let `x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5)` and `y_(1) lt y_(2) lt y_(3) lt y_(4) lt y_(5)` are in AP such that `underset(I = 1)overset(5)(sum) x_(i) = underset(I = 1)overset(5)(sum) y_(i) = 25` and `underset(I =1)overset(5)(II) x_(i) = underset(I = 1)overset(5)(II) y_(i) = 0` then `\|y_(5) - x_(5)\|`A. `beta_(n)" "and log" "beta_(n)`B. `K_(n)" "and log" "beta_(n-1)`C. `beta_(n-1)" "and log" "beta_(n-1)`D. `K_(n-1)" "and log" "K_(n-1)`
Answer» Correct Answer - a `K_(1)K_(2)K_(3) = beta_(3) , beta_(n) = K_(1)K_(2)………..K_(n)` `"log"beta_(n) = "log" K_(1) + "log"K_(2) + "log"K_(3) + ……… + "log"K_(n)`

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