1.

Thesumofnterms of series is 3n^2+ 4n. .Then the series is(a) A.P.(b)G.P.(c) H.P.(d) A.G.S.

Answer»

Sn = 3 n² + 4 nTn = Sn - S_n-1 = 3 n² + 4 n - 3 (n-1)² - 4 (n-1) = 6 n + 1

d = Tn - T_n-1 = 6 n + 1 - 6(n-1) - 1 = 6Since d = 6 = constant, The sequence is an AP

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