First multiple of \(4\) that is greater than \(10\) is \(12\). Next will be \(16\).
Therefore, \(12, 16, 20, 24, ….\)
All these are divisible by \(4\) and thus, all these are terms of an A.P. with first term as \(12\) and common difference is \(4\).
When we divide \(250\) by \(4\), the remainder will be \(2\).
Therefore, \(250 − 2 = 248\) is divisible by \(4\).
The series is as follows.
\(12, 16, 20, 24, …, 248\)
Let \(248\) be the nth term of this A.P.
\(a = 12\)
\(d = 4\)
\(a_n = 248\)
\(a_n = a + (n-1)d\)
\(248 = 12 + (n-1)4\)
\(\frac {236}{4} = n-1\)
\(59 = n-1\)
\(n = 60\)
Therefore, there are \(60\) multiples of \(4\) between \(10\) and \(250\).
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 :