If the ${p^{th}}$ term of an $A.P.$ be $\frac{1}{q}$ and ${q^{th}}$ term be $\frac{1}{p}$, then the sum of its $p{q^{th}}$ terms will be
$\frac{{pq - 1}}{2}$
$\frac{{1 - pq}}{2}$
$\frac{{pq + 1}}{2}$
$ - \frac{{pq + 1}}{2}$
The sum of $1 + 3 + 5 + 7 + .........$ upto $n$ terms is
If $p$ times the ${p^{th}}$ term of an $A.P.$ is equal to $q$ times the ${q^{th}}$ term of an $A.P.$, then ${(p + q)^{th}}$ term is
Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$
The number of terms common to the two A.P.'s $3,7,11, \ldots ., 407$ and $2,9,16, \ldots . .709$ is
The number of terms in an $A .P.$ is even ; the sum of the odd terms in it is $24$ and that the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$ , then the number of terms in the $A.P.$ is