A Level Mathematics

Problem  The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of the 2nd and 3rd terms is 30. Find the sum to infinity. Solution Sometimes, it is better using the term formula and manipulate algebra to obtain a simpler look for the equation. From the given information, we have equation (1) \( a + ar = 50 \implies a(1+r) = 50 \) and equation (2) \( ar + ar^2 = 30 \implies ar(1+r) = 30 \). However, we do know that \(a(1+r) = 50\) and judging from second equation, we have \( r\times 50 = 30 \implies \displaystyle r = \frac{3}{5} \). From here, you can find \( a = \displaystyle \frac{125}{4} \). Once you have a, you can use the formula for sum to infinity \( \displaystyle  S_{\infty} = \frac{a}{1-r} =\frac{625}{8}\)

Back in IGCSE, students should be familiar with this type of expansion \[ (x+2)^2 = (x+2)\times(x+2) = x^2 + 4x + 4\] As you can see from the expansion itself, there are three terms to complete a perfect square a.k.a \((x+2)^2\). Now, if we take a look at similar ones, we can spot the general pattern of this very particular type of expansion \[ \begin{eqnarray*} (x+3)^2 &=& x^2 + 6x + 9 \\ (x+4)^2 &=& x^2 + 8x + 16 \\ (x+6)^2 &=& x^2 + 12x + 36 \\ (x-7)^2 &=& x^2 -14x + 49 \\ \end{eqnarray*} \] That is all the fun with expanding. Now, the real question is: "How can we do the opposite or technically, complete the square?". The answer is simple, you need 3 terms to complete a square. Let's take a look at the following example \[ x^2 + 6x + 11 = (x^2 + 6x + 9) + 2 = (x+3)^2 + 2 \] And it is that simple. Of course, more practice is needed to master this skill. Why is this an important skill? Let's say \( y = x^2 + 6x + 11 ...