0606 P12 - Mar 2025 - Q8 - 7 marks
7190
The first term of a geometric progression is \(10\). This geometric progression has a positive common ratio \(r\).
The first term of an arithmetic progression is also \(10\). This arithmetic progression has a negative common difference \(d\).
The second term of the geometric progression is the same as the fourth term of the arithmetic progression.
The third term of the geometric progression is the same as the sixth term of the arithmetic progression.
(a) Find the values of \(r\) and \(d\).
(b) Determine whether the geometric progression has a sum to infinity.
Solution
Answer: \(r=\frac23\), \(d=-\frac{10}{9}\). The geometric progression has a sum to infinity.
Identify whether the sequence is arithmetic or geometric, then use the correct formula for the required term or sum.
The second term of the geometric progression is \(10r\). The fourth term of the arithmetic progression is \(10+3d\). Hence
\(10r=10+3d.\)
The third term of the geometric progression is \(10r^2\). The sixth term of the arithmetic progression is \(10+5d\). Hence
\(10r^2=10+5d.\)
From the first equation,
\(d=\frac{10r-10}{3}.\)
Substitute this into the second equation:
\(10r^2=10+5\left(\frac{10r-10}{3}\right).\)
Multiplying by \(3\),
\(30r^2=50r-20.\)
So
\(3r^2-5r+2=0.\)
Factorise:
\((3r-2)(r-1)=0.\)
If \(r=1\), then \(d=0\), but \(d\) is negative. Therefore \(r=\frac23\).
Then
\(d=\frac{10\cdot\frac23-10}{3}=-\frac{10}{9}.\)
Since \(|r|=\frac23\lt 1\), the geometric progression has a sum to infinity.
This gives the required answer: \(r=\frac23\), \(d=-\frac{10}{9}\). The geometric progression has a sum to infinity.
