Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals

If $a _{1}, a _{2}, a _{3} \ldots$ and $b _{1}, b _{2}, b _{3} \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1} b_{1}=1=a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to

$8^{th}$ term of the series $2\sqrt 2 + \sqrt 2 + 0 + .....$ will be

Let $a_{1}, a_{2} \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S _{ n }= a _{1}+ a _{2}+\ldots+ a _{ n }$ If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50,$ then the ordered pair $\left( S _{ n -4}, a _{ n -4}\right)$ is equal to

If ${a_1},\;{a_2},\,{a_3},......{a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$, then ${a_1} + {a_2} + {a_3} + ........ + {a_{23}} + {a_{24}} = $

Suppose $a_{1}, a_{2}, \ldots, a_{ n }, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is $5: 17$ and $110< a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -