Please answer no. 12
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The answer is (x+a)A
Given,
A= x1+x2+.....+xn/n
Now adding a to each term and taking average
A(new) = (x+a)x1+(x+a)x2......(x+a)xn/n
As we see (x+a) is common
A(new) = (x+a) x1+x2....xn/n
Thus A(new)=(x+a)A
Given,
A= x1+x2+.....+xn/n
Now adding a to each term and taking average
A(new) = (x+a)x1+(x+a)x2......(x+a)xn/n
As we see (x+a) is common
A(new) = (x+a) x1+x2....xn/n
Thus A(new)=(x+a)A
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