For AP: \(63, 65, 67, ….\)
\(a = 63\) and \(d = a_2 − a_1 = 65 − 63 = 2\)
nth term of this A.P.
\(a_n = a + (n − 1) d \)
\(a_n= 63 + (n − 1) 2 = 63 + 2n − 2 \)
\(a_n = 61 + 2n\ \ \ \ ……(1)\)
For AP: \(3, 10, 17, ….\)
\(a = 3\) and \(d = a_2 − a_1 = 10 − 3 = 7\)
nth term of this A.P. \(= 3 + (n − 1) 7\)
\(a_n = 3 + 7n − 7 \)
\(a_n = 7n − 4 \ \ \ \ ……(2)\)
It is given that, nth term of these A.P.s are equal to each other. Equating both these equations, we obtain
\(61 + 2n = 7n − 4\)
\(61 + 4 = 5n\)
\(5n = 65\)
\(n = 13\)
Therefore, 13th terms of both these A.P.s are equal to each other.
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 :