(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 ...
Question 1 Report
(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 an expression for the nth term.
(b) The equation of a circle is given by \(x^{2} + y^{2} - 10x - 8y + 25 = 0\).
(i) Show that the circle touches the x- axis ; (ii) Find the coordinates of the point of contact.
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: \((5,\,0).\)
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