Given that,
A.P. \(-3, -\frac 12, 2, .......\)
First term \(a = −3\)
Common difference, \(d = a_2 − a_1\)
\(d = -\frac 12 - (-3)\)
\(d = -\frac 12 + 3\)
\(d = \frac {-1+6}{2}\)
\(d = \frac 52\)
We know,
\(a_n = a + (n-1)d\)
\(a_{11} = -3 + (11-1)(\frac 52)\)
\(a_{11} = -3 + 10 \times \frac 52\)
\(a_{11} = -3 + 25\)
\(a_{11}= 22\)
Hence, the correct option is (B): \(22\)
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 :