Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = 1.12+\(\frac{5}{4}=\frac{6}{4}=\frac{3}{2}\)
a2 = 2.22+\(\frac{5}{4}=\frac{2.9}{4}=\frac{9}{2}\)
a3 = 3.32+\(\frac{5}{4}=\frac{3.14}{4}=\frac{21}{2}\)
a4 = 4.42+\(\frac{5}{4}\) = 21
a5 = 5.52+\(\frac{5}{4}=\frac{5.30}{4}=\frac{75}{2}\)
Therefore, the required terms are \(\frac{3}{2},\frac{9}{2},\frac{21}{2},21,and\frac{75}{2}.\)
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