Three numbers are in $A.P.$ such that their sum is $18$ and the sum of their squares is $158$. The greatest number among them is

Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an $A$.$P$. of four terms such that each term of the $A$.$P$. and its common difference $l$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}+l^{4}=361$, then the largest term of the $A$.$P$. is equal to

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 530$ and $S_{5} = 140$,then $S_{20} - S_{6}$ is equal to:

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

Given the sum of the first $n$ terms of an $A.P.$ is $S_n = 2n + 3n^2$. Another $A.P.$ is formed with the same first term and double the common difference. The sum of $n$ terms of the new $A.P.$ is:

If $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$, then $x$ equals