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If anāaa_{n}\to aanāāa as nāān\to\inftynāā and bn=1nāi=1naib_{n}= \frac{1}{n}\sum\limits_{i=1}^{n}a_{i}bnā=n1āi=1ānāaiā, then bnāa\mboxasnāāb_{n}\to a\mbox{ as }n\to\inftybnāāa\mboxasnāā