Answer:
[tex]x-n+1[/tex]
Step-by-step explanation:
Given the mean of [tex]n[/tex] terms as [tex]x[/tex], and of [tex]n-1[/tex] as [tex]x+1[/tex]:
We know that [tex]\bar X=\frac{1}{n}\sum x_i[/tex] for [tex]n[/tex] terms.
Therefore [tex]\sum x_n=n \bar X[/tex] for [tex]n[/tex] terms;
For [tex]n-1[/tex] terms; [tex]\sum x_n_-_1=(n-1)(x+1)=nx-x+n-1[/tex]
To find the nth score, we subtract the sum of (n-1) terms from the sum of n terms:
[tex]n^t^h[/tex] term=
[tex]nx-(nx-x+n-1)\\=x-n+1[/tex]
