Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = 2×1-\(\frac{3}{6} =\frac{-1}{6}\)
a2 = 2×2-\(\frac{3}{6} = \frac{1}{6}\)
a3 = 2×3-\(\frac{3}{6} = \frac{3}{6} = \frac{1}{2}\)
a4 = 2×4-\(\frac{3}{6} = \frac{5}{6}\)
a5 = 2×5-\(\frac{3}{6} = \frac{7}{6}\)
Therefore, the required terms are \(\frac{-1}{6}, \frac{1}{6},\frac{1}{2}, \frac{5}{6}, and \frac{7}{6}\)
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to