\(3, 8, 13, 18, ….\)
For this A.P., \(a = 3\) and \(d = a_2 − a_1 = 8 − 3 = 5\)
Let nth term of this A.P. be 78.
\(a_n = a + (n − 1) d \)
\(78 = 3 + (n − 1) 5\)
\(75 = (n − 1) 5\)
\((n − 1) = 15\)
\(n = 16\)
Hence, \(16^{th}\) term of this A.P. is 78.
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 :