If $19^{th}$ terms of non -zero $A.P.$ is zero, then its ($49^{th}$ term) : ($29^{th}$ term) is
$4 : 1$
$1 : 3$
$3 : 1$
$2 : 1$
What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?
Let $S_n$ denote the sum of the first $n$ terms of an $A.P$.. If $S_4 = 16$ and $S_6 = -48$, then $S_{10}$ is equal to
The number of $5 -$tuples $(a, b, c, d, e)$ of positive integers such that
$I.$ $a, b, c, d, e$ are the measures of angles of a convex pentagon in degrees
$II$. $a \leq b \leq c \leq d \leq e$
$III.$ $a, b, c, d, e$ are in arithmetic progression is
Find the $20^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=\frac{n(n-2)}{n+3}$
If $\log _{10} 2, \log _{10} (2^x + 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-