Ans: C
Let $T(n) = 2n-19$, where $n$ is a positive integer.
Let $T(n) = 2n-19$, where $n$ is a positive integer.
I is true. $T(22)=2(22)-19=25$. Therefore, $25$ is a term of the sequence.
II is false. Let $T(k)$ be the first positive term of the sequence.
$\begin{array}{rcl}
T(k) & > & 0 \\
2k-19 & > & 0 \\
2k & > & 19 \\
k & > & 9.5
\end{array}$
Therefore $T(10)$ is the first positive term of the sequence. Hence, the sequence has $9$ negative terms.
III is true. Note that $T(1) = 2(1)-19 = -17$ and $T(2) = 2(2) – 19 =-15$. Therefore the common difference is $2$.
Hence, we have
$\begin{array}{rcl}
S(n) & = & \dfrac{n}{2} [ 2(-17) + (n-1)(2) ] \\
& = & n(n-18) \\
& = & n^2 -18n
\end{array}$
