If mean of the series \(x_1\), \(x^2\), ….. , \(x_n\) is \(\bar{x}\), then the mean of the series \(x_i\) + 2i, i = 1, 2, ……, n will be

Solution :

As given \(\bar{x}\) = \(x_1 + x_2 + …. + x_n\over n\)

If the mean of the series \(x_i\) + 2i, i = 1, 2, ….., n be \(\bar{X}\), then

\(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\over n\)

= \(x_1 + x_2 + …. + x_n\over n\) + \(2(1+2+3+….+n)\over n\)

= \(\bar{x}\) + \(2n(n+1)\over 2n\)

= \(\bar{x}\) + n + 1.


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