It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs. \(200\).
Therefore, the salaries of each year after 1995 are:
\(5000, 5200, 5400, ……\)
Here,
\(a = 5000\) and \(d = 200\)
Let after \(n^{th}\) year, his salary be Rs. \(7000\).
Therefore,
\(a_n = a + (n − 1) d\)
\(7000 = 5000 + (n − 1) 200\)
\(200(n − 1) = 2000\)
\((n − 1) = 10\)
\(n = 11\)
Therefore, in \(11^{th}\) year, his salary will be Rs. \(7000\).
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 :