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If an = \(\cfrac{n(n+3)}{n+2}\) , then find a17 = ……………A) \(\cfrac{340}{20}\)B) \(\cfrac{341}{19}\)C) \(\cfrac{340}{19}\)D) \(\cfrac{341}{20}\) |
Answer» Correct option is (C) \(\frac{340}{19}\) We have \(a_n = \frac{n(n+3)}{n+2}\) \(\therefore\) \(a_{17}=\frac{17(17+3)}{17+2}\) (Putting n = 17) \(=\frac{17\times20}{19}=\frac{340}{19}\) Correct option is C) \(\cfrac{340}{19}\) |
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