(a) \(S_{n} = \frac{5n^{2}}{2} + \frac{5n}{2}\)
\(S_{1} = \frac{5(1^{2})}{2} + \frac{5(1)}{2} = 5 \implies T_{1} = 5\)
\(S_{2} = \frac{5(2^{2})}{2} + \frac{5(2)}{2} = 10 + 5 = 15\)
\(S_{3} = \frac{5(3^{2})}{2} + \frac{5(3)}{2} = 30\)
\(S_{4} = \frac{5(4^{2})}{2} + \frac{5(4)}{2} = 50\)
\(T_{2} = S_{2} - S_{1} = 15 - 5 = 10\)
\(T_{3} = S_{3} - S_{2} = 30 - 15 = 15\)
\(T_{4} = 50 - 30 = 20\)
The first four terms are 5, 10, 15 and 20.
This is a linear sequence with d = 5.
\(T_{n} = a + (n - 1)d\)
= \(5 + (n - 1)5 \implies T_{n} = 5 + 5n - 5 = 5n\)
(b) \(x^{2} + y^{2} - 10x - 8y + 25 = 0\)
For circle to touch x- axis, the distance from centre of circle to the x- axis must be equal to the radius.
\(x^{2} + y^{2} - 10x - 8y + 25 = 0\)
\(2g = -10, g = -5\)
\(2f = -8 , f = -4\)
Centre = (-g, -f) = (5, 4).
\(Radius = \sqrt{g^{2} + f^{2} - c} = \sqrt{(-5)^{2} + (-4)^{2} - 25} = 4\)

\(|PQ| = \sqrt{4^{2}} = 4\)
Since PQ = radius, circle touches x- axis.