This is a geometric series with the first term \( a = 1 \) and the common ratio \( r = \frac{1}{3} \).
The sum of an infinite geometric series is given by the formula: \[ S = \frac{a}{1 - r}. \]
Substituting \( a = 1 \) and \( r = \frac{1}{3} \): \[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}. \]
Thus, the sum of the series is \( \frac{3}{2} \).
Let $a_1, a_2, \ldots, a_n$ be in AP If $a_5=2 a_7$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to
Let $a_1, a_2, a_3, \ldots$ be an AP If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to :
Match List-I with List-II: List-I