Find the values of n and in each of the following cases:

(i) = -10 and = 62


(ii) = 30 and = 150

(i) Given, = -10

= (x1 – 12) + (x2 – 12) + …. + (xn + 12) = -10


= (x1 + x2 + x3 + x4 + ….. + xn) + (12 + 12 + 12 + …. + 12) = -10


= ∑x – 12n = -10 (I)


And = 62


= (x1 – 3) + (x2 – 3) + (x3 – 3) + …. + (xn – 3) = 62


= (x1 + x2 + x3 + …. + xn) – (3 + 3 + 3 + …. + 3) = 62


= ∑x – 3n = 62 (II)


BY subtracting (I) from (II), we get


∑x – 3n - ∑x – 12n = 62 + 10


= 9n = 72


= n =


= 8


Put value of n in equation (I), we get


∑x – 12 * 8 = -10


= ∑x – 96 = -10


= ∑x = -10 + 96 = 86


Therefore, =


=


= 10.75


(ii) Given, = 30


= (x1 – 10) + (x2 – 10) + …. + (xn – 10) = 30


= (x1 + x2 + x3 + …. + xn) – (10 + 10 + 10 + …. + 10) = 30


= ∑x – 10n = 30 (I)


And = 150


= (x1 – 6) + (x2 – 6)+ …. + (xn – 6) = 150


= (x1 + x2 + x3 + …. + xn) – (6 + 6 + 6 + …. + 6) = 150


= ∑x – 6n = 150


By subtracting (I) from (II), we get


∑x – 6n - ∑x – 10n = 150 – 30


= ∑x - ∑x + 4n = 120


n =


= 30


Put value of n in (I), we get


∑x – 10 * 30 = 30


∑x – 300 = 30


∑x = 30 + 300


= 330


Therefore, =


=


= 11


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