2018-II-35

Ans: B
I is not true.

$\begin{array}{rcl}
T(n) & = & S(n) – S(n-1) \\
T(n) & = & (6n^2 -n) – [6(n – 1)^2 -(n – 1)]\\
T(n) & = & 6n^2 – n – 6n^2 +12n – 6 + n -1\\
T(n) & = & 12n – 7 \\
\end{array}$

For some $k$ such that $T(k) = 22$,

$\begin{array}{rcl}
12k – 7 & = & 22 \\
12 k & = & 29 \\
k & = & \dfrac{29}{12}
\end{array}$

Since $k$ must be an positive integer, then $22$ is not a term of the sequence.

II is true.

$\begin{array}{rcl}
T(1) & = & S(1) \\
T(1) & = & 6(1)^2 – 1 \\
T(1) & = & 5
\end{array}$

III is not true.

$\begin{array}{rcl}
\dfrac{T(n)}{T(n-1)} & = & \dfrac{12n-7}{(12(n-1) – 7} \\
& = & \dfrac{12n – 7}{12n – 19} \text{, which is not a constant.}
\end{array}$