Calculate the common difference \((d)\):

\begin{align*} a & = -\frac{1}{2} \\ \text{General formula: } T_{n} &= a + (n-1)d \\ T_{22} &= -\frac{1}{2} + (22 - 1)d \\ \therefore 10 &= -\frac{1}{2} + 21d \\ 10 + \frac{1}{2} &= 21d \\ \frac{21}{2} &= 21d \\ \therefore d &= \frac{1}{2} \end{align*}

Now determine the general term for the sequence:

\begin{align*} T_{n} &= a + (n - 1)d \\ T_{n} &= -\frac{1}{2} + (n-1) \left( \frac{1}{2} \right)\\ &= -\frac{1}{2} + \frac{1}{2}n -\frac{1}{2} \\ \therefore T_{n} &= \frac{1}{2}n - 1 \end{align*}

• There is a common difference between any two successive terms in the sequence.
• The graph of \(T_{n}\) vs. \(n\) is a straight line.

• Use the given terms to calculate the common difference \((d): \quad d = T_{2} - T_{1}\).
• From given terms, we know that \(T_{1} = a\).
• Substitute the values for \(a\) and \(d\) into the equation \(T_{n} = a + (n-1)d\).
• Simplify and gather like \(n\) terms.

\begin{align*} d &= T_{2} - T_{1} \\ &= (2p+2) -(p) \\ &= p + 2\\ \text{ or } & \\ d &= T_{3} - T_{2} \\ &= (5p+3) - (2p+2) \\ &= 3p + 1 \\ \therefore 3p + 1 &= p+2 \\ 2p &= 1 \\ \therefore p &= \frac{1}{2} \end{align*} \begin{align*} T_{15} &= a + 14d \\ &= p +14(p + 2) \\ &= 15p + 28 \\ &= 15 \left( \frac{1}{2} \right) + 28 \\ &= 35\frac{1}{2} \end{align*}

\begin{align*} \frac{3a - 2 + x}{2} &= 4a - 4 \\ 3a - 2 + x &= 8a - 8 \\ \therefore x &= 5a - 6 \\ \text{ or }& \\ T_{2} - T_{1} &= T_{3} - T_{2} \\ \therefore (4a - 4) - (3a - 2) &= (x) - (4a - 4) \\ 4a - 4 -3a + 2 &= x -4a + 4 \\ \therefore 5a - 6 &= x \end{align*}

Let the arithmetic means be:

\(a_{1}; a_{2}; a_{3}; a_{4}; a_{5}; a_{6}; a_{7}\)

Therefore the sequence is:

\[T_{1}; a_{1}; a_{2}; a_{3}; a_{4}; a_{5}; a_{6}; a_{7}; T_{9}\] \begin{align*} T_{1} &= 3s-t \\ T_{9} &= -13s + 7t \\ d &= \frac{T_{9} - T_{1}}{8} \\ &= \frac{-13s + 7t -3s + t}{8} \\ &= \frac{-16s + 8t}{8} \\ &= -2s + t \\ \therefore T_{1} &= 3s -t \\ a_{1} &= (3s - t) +(-2s + t) \\ &= s \\ a_{2} &= (s) +(-2s + t) \\ &= -s + t \\ a_{3} &= (-s + t) +(-2s + t) \\ &= -3s + 2t \\ a_{4} &= (-3s + 2t) +(-2s + t) \\ &= -5s + 3t \\ a_{5} &= (-5s + 3t) +(-2s + t) \\ &= -7s + 4t \\ a_{6} &= (-7s + 4t) +(-2s + t) \\ &= -9s + 5t \\ a_{7} &= (-9s + 5t) +(-2s + t) \\ &= -11s + 6t \\ T_{9} &= (-11s + 6t) +(-2s + t)\\ &= -13s + 7t \end{align*}

Starting with the first term, we have \(n = 1\) and \(T_{1} = -1\): \begin{align*} T_{1} & = (1)^{2} + b(1) + c \\ (-1) & = 1 + b + c \\ -2 & = b + c \end{align*}

For the second term, we use \(n = 2\) and \(T_{2} = 2\):

\begin{align*} T_{2} & = (2)^{2} + b(2) + c \\ (2) & = 4 + 2 b + c \\ -2 & = 2 b + c \end{align*}

Now we must solve these equations simultaneously. We can do this by substitution, but here we will show the solution using the 'elmination' method (which means subtracting one equation from the other to cancel the \(c\)'s).

\begin{align*} -2 & = 2 b + c \\ - ( -2 & = \; b + c ) \\ 0 & = \; b \end{align*}

Finally, calculate the value of \(c\). As usual for simultaneous equations, this means that we must substitute the \(b = 0\) into either of the equations we used above. Let's use the equation \(-2 = b + c\).

\begin{align*} b = 0 \; \longrightarrow \quad -2 & = b + c \\ -2 & = ( 0 ) + c \\ -2 & = \; c \end{align*}

The final answers are \(b = 0\) and \(c = -2\).

NOTE: Now we know that the general term of the sequence is \(\; T_{n} = n^{2} - 2 \;\). We can use this to check our answers. We know that \(T_{3} = 7\). Substitute \(n = 3\) into the general formula to check:

\begin{align*} T_{n} & = n^{2} - 2 \\ T_{3} & = (3)^{2} - 2 \\ & = (9) +0 -2 \\ & = 7 \end{align*}

Starting with the first term, we have \(n = 1\) and \(T_{1} = 2\): \begin{align*} T_{1} & = (1)^{2} + b(1) + c \\ (2) & = 1 + b + c \\ 1 & = b + c \end{align*}

For the second term, we use \(n = 2\) and \(T_{2} = 7\): \begin{align*} T_{2} & = (2)^{2} + b(2) + c \\ (7) & = 4 + 2 b + c \\ 3 & = 2 b + c \end{align*}

Now we must solve these equations simultaneously. We can do this by substitution, but here we will show the solution using the 'elimination' method (which means subtracting one equation from the other to cancel the \(c\)'s). \begin{align*} 3 & = 2 b + c \\ - ( 1 & = \; b + c ) \\ 2 & = \; b \end{align*}

Finally, calculate the value of \(c\). As usual for simultaneous equations, this means that we must substitute the \(b = 2\) into either of the equations we used above. Let's use the equation \(1 = b + c\). \begin{align*} b = 2 \; \longrightarrow \quad 1 & = b + c \\ 1 & = ( 2 ) + c \\ -1 & = \; c \end{align*} The final answers are \(b = 2\) and \(c = -1\).

NOTE: Now we know that the general term of the sequence is \(\; T_{n} = n^{2} + 2 n - 1 \;\). We can use this to check our answers. We know that \(T_{3} = 14\). Substitute \(n = 3\) into the general formula to check:

\begin{align*} T_{n} & = n^{2} + 2 n - 1 \\ T_{3} & = (3)^{2} + 2 (3) - 1 \\ & = (9) +6 -1 \\ & = 14 \end{align*}

\begin{align*} \text{Let } T_2 &= x \\ \therefore T_2 - T_1 &= x - 4 \\ \text{And } T_3 - T_2 &= 34 - x \\ \text{Second difference } &= (T_3 - T_2) - (T_2 - T_1) \\ &= (34 - x) - (x - 4)\\ \therefore 10 &= 38 - 2x \\ 2x &= 28 \\ \therefore x &= 14 \end{align*}

\(4; 14; 34; 64; 104; 154\)