(a) The sum of the first n terms of a sequence is given by \(S_{n} = \frac{5n^{2}}{2} + \frac{5n}{2}\). Write down the first four terms of the sequence and ...

Answer Details

(a) Sequence with \(S_n=\dfrac{5n^2}{2}+\dfrac{5n}{2}=\dfrac{5n(n+1)}{2}.\)

Each term is \(T_n=S_n-S_{n-1}.\) Compute partial sums:

\(S_1=\dfrac{5(1)(2)}{2}=5,\ S_2=\dfrac{5(2)(3)}{2}=15,\ S_3=\dfrac{5(3)(4)}{2}=30,\ S_4=\dfrac{5(4)(5)}{2}=50.\)

Terms: \(T_1=5,\ T_2=15-5=10,\ T_3=30-15=15,\ T_4=50-30=20.\)

First four terms: 5, 10, 15, 20.

General term:

\[T_n=S_n-S_{n-1}=\frac{5}{2}\big[n(n+1)-(n-1)n\big]=\frac{5}{2}\,n\,[(n+1)-(n-1)]=\frac{5}{2}\,n(2)=5n.\]

So \(T_n=5n.\)

(b) Circle \(x^2+y^2-10x-8y+25=0.\)

Complete the square:

\[(x^2-10x)+(y^2-8y)+25=0\Rightarrow (x-5)^2-25+(y-4)^2-16+25=0,\]

\[(x-5)^2+(y-4)^2=16.\]

Centre \((5,4),\) radius \(r=4.\)

(i) Touching the x-axis. The perpendicular distance from the centre \((5,4)\) to the x-axis \((y=0)\) is \(4,\) which equals the radius. Since distance \(=\) radius, the circle is tangent to (touches) the x-axis.

(ii) Point of contact. The contact point is the foot of the perpendicular from the centre to the x-axis, i.e. directly below the centre: \((5,\,0).\)