For this A.P., \(a = 11\) and \(d = a_2 − a_1 = 8 − 11 = −3\)
Let \(−150\) be the nth term of this A.P.
We know that,
\(a_n = a + (n-1)d\)
\(-150 = 11 + (n-1)(-3)\)
\(-150 = 11-3n+3\)
\(-164 = -3n\)
\(n = \frac {164}{3}\)
Clearly, n is not an integer.
Therefore, \(−150\) is not a term of this A.P.
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 :