If $a_1, a_2, a_3, …….$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is
$200$
$280$
$150$
$120$
What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?
If the sum of the $10$ terms of an $A.P.$ is $4$ times to the sum of its $5$ terms, then the ratio of first term and common difference is
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
The first term of an $A.P.$ of consecutive integers is ${p^2} + 1$ The sum of $(2p + 1)$ terms of this series can be expressed as
If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$, then their common difference will be