The sixth term of an $A.P.$ is equal to $2$, the value of the common difference of the $A.P.$ which makes the product ${a_1}{a_4}{a_5}$ least is given by
Find the $9^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=(-1)^{n-1} n^{3}$
For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
If the sum of the first $2n$ terms of $2,\,5,\,8...$ is equal to the sum of the first $n$ terms of $57,\,59,\,61...$, then $n$ is equal to
If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are