Find the values of n and `bar X` in each of the following cases: (i) `sum_(i=1)^n (x_1-12)=-10` and `sum_(i=1)^n (x_1-3)=62` (ii) `sum_(i=1)^n (x_1-10)=30` and `sum_(i=1)^n (x_1-6)=150`

Explanation

`(1)` Given
`sum_(i=1)^n(x_1-12)=-10`
`=(x1 −12) + (x2 – 12) = +(xn - 12) = - 10`
`=(x1+ x2 + x3 + x4+x5 + +xn) (12+12 + 12......... + 12) = -10`
`sum_x - 12n = -10n`.........`(1)`
And,
`sum_(i=1)^n (x_1-3)=62`
`=(x1-3) + (x2-x3) + (x3 − 3) + .............+(xn-3)=62`.
`(x1 + x2 + +xn)-(3+3+3+3+...+37)=62`
`sum(x) - 3n = 62`..........`(2)`
By subtracting equation `(1)` from equation `(2)` We get
`sum(x) - 3n - sum(x) -12n = 62 + 10`
`= 9n=72`
`n=72/9`
`= 8`
putting value of eq `(1)` we get
`sum(x) -12( xx 8) = -10`
`= sum(x) - 96 = 10`
`= sum(x) = - 10 + 96 = 86`
Therefore, `bar x = sum(x)/n`
`=86/8`
`= 10.75`
(ii) Given `sum_(i=1)^n (x_1-10)=30`
`= (x1 - 10) + (x2 - 10) +.......+ (xn - 10) = 30`
`(x1 + x2 + x3 +.......+ Xn)- (10 + 10 + 10 + ... + 10) = 30`
`sum(x) - 10n = 30`........`(1)`
And
`sum_(i=1)^n (x_1-6)=150`
`= (x1 - 6)+ (x2 − 6)+ ....... + (xn - 6) = 150`
`= (x1 + x2 + x3 +...... + Xn)- (6+6+6+ ... + 6) = 150`
by subtracting `(1)` from`(2)` we get
`sum(x) - 6n - sum(x) -10n = 150 - 30`
`sum(x) - sum(x) + 4n = 120`
`n= 120/4`
` =30`
Put value of n in `(1)`, we get
`sum(x) - 10(xx30) = 30`
`sum(x) - 300 = 30`
`sum(x) = 30 + 300 = 330`
Therefore, `bar x = sum(x)/n`
`= 330/30`
`=11`
Hence the required answer is `11`