Common ratio, r = 2
Let a be the first term of the G.P.
∴ a8 = ar 8–1 = ar7
⇒ ar7 = 192
a(2) 7 = 192
a(2) 7 = (2)6 (3)
⇒a = (2)6 × \(\frac{3}{(2)}7=\frac{3}{2}\)
∴ a12 = ar12 - 1 = \((\frac{3}{2})\)(2)11 = (3)(2)10 = 3072
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