After inserting $n$, $A.M.'s$ between $2$ and $38$, the sum of the resulting progression is $200$. The value of $n$ is
$10$
$8$
$9$
None of these
What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?
Let $a_n, n \geq 1$, be an arithmetic progression with first term $2$ and common difference $4$ . Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum \limits_{n=1}^{10} M_n$ is
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
Let ${S_1},{S_2},......,{S_{101}}$ be the consecutive terms of an $A.P$ . If $\frac{1}{{{S_1}{S_2}}} + \frac{1}{{{S_2}{S_3}}} + .... + \frac{1}{{{S_{100}}{S_{101}}}} = \frac{1}{6}$ and ${S_1} + {S_{101}} = 50$ , then $\left| {{S_1} - {S_{101}}} \right|$ is equal to
Write the first five terms of the following sequence and obtain the corresponding series :
$a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n\,>\,2$