Standard +0.3 This is a straightforward system-of-equations problem involving standard AP and GP formulas. Given the first term (3), students set up two equations using a₃ = 3 + 2d and a₁₃ = 3 + 12d for the GP terms, then solve simultaneously. It requires careful algebraic manipulation but follows a familiar pattern with no conceptual surprises, making it slightly easier than average.
4 The 1st, 3rd and 13th terms of an arithmetic progression are also the 1st, 2nd and 3rd terms respectively of a geometric progression. The first term of each progression is 3 . Find the common difference of the arithmetic progression and the common ratio of the geometric progression.
\(r=\frac{3+2d}{3}\) or \(\frac{3+12d}{3+2d}\) or \(r^2=\frac{3+12d}{3}\)
B1
1 correct equation in \(r\) and \(d\) only is sufficient
\((3+2d)^2=3(3+12d)\) oe OR sub \(2d=3r-3\)
M1
Eliminate \(r\) or \(d\) using valid method
\((4)d(d-6)=0\) OR \(3r^2=18r-15\rightarrow(r-1)(r-5)\)
DM1
Attempt to simplify and solve quadratic
\(d=6\)
A1
Ignore \(d=0\) or \(r=1\)
\(r=5\)
A1 [5]
Do not allow \(-5\) or \(\pm 5\)
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r=\frac{3+2d}{3}$ or $\frac{3+12d}{3+2d}$ or $r^2=\frac{3+12d}{3}$ | B1 | 1 correct equation in $r$ and $d$ only is sufficient |
| $(3+2d)^2=3(3+12d)$ oe OR sub $2d=3r-3$ | M1 | Eliminate $r$ or $d$ using valid method |
| $(4)d(d-6)=0$ OR $3r^2=18r-15\rightarrow(r-1)(r-5)$ | DM1 | Attempt to simplify and solve quadratic |
| $d=6$ | A1 | Ignore $d=0$ or $r=1$ |
| $r=5$ | A1 [5] | Do not allow $-5$ or $\pm 5$ |
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4 The 1st, 3rd and 13th terms of an arithmetic progression are also the 1st, 2nd and 3rd terms respectively of a geometric progression. The first term of each progression is 3 . Find the common difference of the arithmetic progression and the common ratio of the geometric progression.
\hfill \mbox{\textit{CAIE P1 2016 Q4 [5]}}