Moderate -0.3 This is a straightforward geometric progression question requiring recall of standard formulas (ar = 24, a/(1-r) = 150) and solving simultaneous equations. The algebra is routine and the question is slightly easier than average as it explicitly tells students to write down the equations first, providing scaffolding for a standard textbook exercise.
The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in \(a\) and \(r\), where \(a\) is the first term and \(r\) is the common ratio. Solve your equations to find the possible values of \(a\) and \(r\). [5]
Question 4:
4 | ar = 24 (i)
a
150 (ii)
1 r
correct substitution to eliminate one
unknown
r = 0.8 or 0.2
a = 30 or a = 120 | B1*
B1*
M1dep*
A1
A1
[5] | 150 a
eg subst. of a = 150(1 – r) or r
150
in (i)
24 24
alternatively, subst. of a or r
r a
in (ii)
or A1 for each correct pair of values
ignore incorrect pairing if correct values
already correctly attributed | allow ar2 – 1 = 24
if M0, B1 for both values of r and
B1 for both values of a, or B1 for
each pair of correct values
NB 150r2 – 150r + 24 [= 0]
a2 – 150a +3600 [= 0]
A0 if wrongly attributed
A0 if wrongly attributed
The second term of a geometric progression is 24. The sum to infinity of this progression is 150. Write down two equations in $a$ and $r$, where $a$ is the first term and $r$ is the common ratio. Solve your equations to find the possible values of $a$ and $r$. [5]
\hfill \mbox{\textit{OCR MEI C2 Q4 [5]}}