Given that,
\(a = 5\), \(d = 1.75\) and \(a_n = 20.75\).
\(n = ?\)
\(a_n = a + (n − 1) d\)
\(20.75 = 5 + (n-1)1.75\)
\(15.75 = (n-1)1.75\)
\(n-1 = \frac {15.75}{1.75}\)
\(n-1 = \frac {1575}{175}\)
\(n-1 = \frac {63}{7}\)
\(n − 1 = 9\)
\(n = 10\)
Hence, n is 10.
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 :