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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