Standard +0.3 This is a straightforward geometric progression problem requiring students to use the common ratio property (r = T₂/T₁ = T₃/T₂) to form and solve a quadratic equation for p, then find r and apply the sum to infinity formula. While it involves multiple steps, each is standard: forming the equation (-2p)²=(2p+6)(p+2), solving the quadratic, checking |r|<1, and applying S∞=a/(1-r). This is slightly easier than average as it follows a well-practiced procedure with no conceptual surprises.
2 The first, second and third terms of a geometric progression are \(2 p + 6 , - 2 p\) and \(p + 2\) respectively, where \(p\) is positive.
Find the sum to infinity of the progression.
Correct formula used with their values for \(a\) and \(r\), \(
\((s_\infty =)\ 10.8\)
A1
OE. A0 if an extra solution given. SC B2 for \(s_\infty = \frac{2p+6}{1-\frac{-2p}{2p+6}}\) or \(\frac{2p+6}{1-\frac{p+2}{-2p}}\) ignore any subsequent algebraic simplification.
## Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $(-2p)^2 = (2p+6)(p+2)$ or $\frac{-2p}{2p+6} = \frac{p+2}{-2p}$ | M1 | OE. Using "$a, b, c$ then $b^2 = ac$" or $a = 2p+6$, $ar = -2p$ and $ar^2 = p+2$ to form a correct relationship in terms of $p$ only |
| $(2p^2 - 10p - 12 = 0)\ p = 6$ | A1 | |
| $a = 18$ and $r = -\frac{2}{3}$ | A1 | |
| $(s_\infty) = \text{their } a \div (1 - \text{their } r)$ $\left(= 18 \div \frac{5}{3}\right)$ | M1 | Correct formula used with their values for $a$ and $r$, $|r| < 1$. Both $a$ & $r$ from the same value of $p$. |
| $(s_\infty =)\ 10.8$ | A1 | OE. A0 if an extra solution given. **SC B2** for $s_\infty = \frac{2p+6}{1-\frac{-2p}{2p+6}}$ or $\frac{2p+6}{1-\frac{p+2}{-2p}}$ ignore any subsequent algebraic simplification. |
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2 The first, second and third terms of a geometric progression are $2 p + 6 , - 2 p$ and $p + 2$ respectively, where $p$ is positive.
Find the sum to infinity of the progression.\\
\hfill \mbox{\textit{CAIE P1 2020 Q2 [5]}}