We know that, For an A.P.,
\(a_n = a + (n − 1) d\)
\(a_{17} = a + (17 − 1) d\)
\(a_{17} = a + 16d\) ……. (i)
Similarly,
\(a_{10} = a + (10− 1) d\)
\(a_{10}= a + 9d\) ……. (ii)
It is given that,
\(a_{17} − a_{10} = 7\)
\((a + 16d) − (a + 9d) = 7\)
\(7d = 7\)
\(d = 1\)
Therefore, the common difference is 1.
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 :