Given that,
A.P. \(10, 7, 4, .....\)
First term, \(a = 10 \)
Common difference, \(d = a_2 − a_1 = 7 − 10 = −3\)
We know that,
\(a_n = a + (n − 1) d \)
\(a_{30} = 10 + (30 − 1) (−3)\)
\(a_{30} = 10 + (29) (−3)\)
\(a_{30}= 10 − 87\)
\(a_{30}= −77\)
Hence, the correct option is (C): \(-77\)
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 :