\begin{align*} T_1 & = S_1 \\ & = 2(1)^2 + 5(1) \\ & = 7 \\ \\ S_2 & = 2(2)^2 + 5(2) \\ & = 18 \\ \\ T_2 & = S_2 - S_1 \phantom{000000} [T_n = S_n - S_{n - 1}] \\ & = 18 - 7 \\ & = 11 \\ \\ S_3 & = 2(3)^2 + 5(3) \\ & = 33 \\ \\ T_3 & = S_3 - S_2 \\ & = 33 - 18 \\ & = 15 \end{align*}

\begin{align*} \text{Common difference, } d & = 11 - 7 \\ & = 4 \\ \\ T_n & = T_1 + (n - 1)(d) \\ & = 7 + (n - 1)(4) \\ & = 7 + 4n - 4 \\ & = 4n + 3 \end{align*}

\begin{align*} \text{Let } T_n & = 420 \\ 4n + 3 & = 420 \\ 4n & = 420 - 3 \\ 4n & = 417 \\ n & = {417 \over 4} \\ n & = 104.25 \\ \\ \text{Since } n \text{ is not an integer, } 420 & \text{ is not a term in the sequence} \end{align*}

\begin{align*} 1^2, & \phantom{.} 2^2, 3^2, \ldots \\ \\ n \text{th term} & = n^2 \\ \\ \\ 5, & \phantom{.} 8, 11, \dots \\ \\ \text{Common difference, } d & = 8 - 5 \\ & = 3 \\ \\ n \text{th term} & = T_1 + (n - 1)(d) \\ & = 5 + (n - 1)(3) \\ & = 5 + 3n - 3 \\ & = 3n + 2 \\ \\ \\ \therefore T_n & = n^2 + 3n + 2 \end{align*}

\begin{align*} \text{Let } T_n & = 132 \\ n^2 + 3n + 2 & = 132 \\ n^2 + 3n + 2 - 132 & = 0 \\ n^2 + 3n - 130 & = 0 \phantom{000000} [\text{Quadratic equation}] \\ (n - 10)(n + 13) & = 0 \end{align*} \begin{align*} n - 10 & = 0 && \text{ or } & n + 13 & = 0 \\ n & = 10 &&& n & = -13 \text{ (N.A.)} \end{align*} $$ \therefore \text{132 is the 10th term.} $$

\begin{align*} T_1 & = 1^2 + 5 = 1 + 5 \\ \\ T_2 & = 2^2 + 8 = 4 + 8 \\ \\ T_3 & = 3^2 + 11 = 9 + 11 \\ \\ T_4 & = 4^2 + 14 = 16 + 14 \\ \\ \text{The sequence alternates } & \text{between the sum of two odd numbers and} \\ \text{the sum of two even num} & \text{bers. The sum is always even.} \end{align*}

\begin{align*} T_n & = n^2 + 3n + 2 \\ & = (n + 1)(n + 2) \\ \\ (n + 1) \text{ and } (n + 2) & \text{ are consecutive integers and one of them is even} \\ \\ \text{The product of an } & \text{odd integer and an even integer will always be even} \\ \\ \therefore \text{Every term in the } & \text{sequence is an even integer.} \end{align*}

\begin{align*} \text{Numerators: } & 4,\ 7,\ 10,\ 13,\ \ldots \\ \\ \text{Common difference, } d & = 7 - 4 \\ & = 3 \\ \\ n \text{th term} & = T_1 + (n - 1)(d) \\ & = 4 + (n - 1)(3) \\ & = 4 + 3n - 3 \\ & = 3n + 1 \\ \\ \\ \text{Denominators: } & 3,\ 7,\ 11,\ 15,\ \ldots \\ \\ \text{Common difference, } d & = 7 - 3 \\ & = 4 \\ \\ n \text{th term of denominator} & = T_1 + (n - 1)(d) \\ & = 3 + (n - 1)(4) \\ & = 3 + 4n - 4 \\ & = 4n - 1 \\ \\ \\ \therefore T_n & = {3n + 1 \over 4n - 1} \end{align*}

\begin{align*} T_n & = {3n + 1 \over 4n - 1} \\ \\ \text{Denominator} - \text{Numerator} & = 4n - 1 - (3n + 1) \\ & = 4n - 1 - 3n - 1 \\ & = n - 2 \\ \\ \text{For } n & \ge 3, \\ n - 2 & \ge 3 - 2 \\ n - 2 & \ge 1 \\ \\ \implies \text{Denominator} & > \text{Numerator for } n \ge 3 \\ \\ \therefore T_n \text{ is less than } & 1 \text{ for } n \ge 3 \end{align*}