0606 P22 - Nov 2021 - Q10 - 10 marks
8062
(a) The first three terms of an arithmetic progression are \(x\), \(5x-4\) and \(8x+2\). Find \(x\) and the common difference.
(b) The first three terms of a geometric progression are \(y\), \(5y-4\) and \(8y+2\).
(i) Find the two possible values of \(y\).
(ii) For each of these values of \(y\), find the corresponding value of the common ratio.
Solution
Answer: \(x=10,\ d=36\); \(y=\frac8{17}\) gives \(r=-\frac72\), and \(y=2\) gives \(r=3\).
(a) In an arithmetic progression, consecutive differences are equal. Therefore
\((5x-4)-x=(8x+2)-(5x-4).\)
So
\(4x-4=3x+6.\)
Hence
\(x=10.\)
The common difference is
\(d=(5x-4)-x=4x-4.\)
Substitute \(x=10\):
\(d=40-4=36.\)
(b)(i) In a geometric progression, consecutive ratios are equal, so
\(\frac{5y-4}{y}=\frac{8y+2}{5y-4}.\)
Cross-multiply:
\((5y-4)^2=y(8y+2).\)
Expand:
\(25y^2-40y+16=8y^2+2y.\)
Rearrange:
\(17y^2-42y+16=0.\)
Factorise:
\((17y-8)(y-2)=0.\)
Therefore
\(y=\frac8{17}\quad\text{or}\quad y=2.\)
(b)(ii) The common ratio is
\(r=\frac{5y-4}{y}.\)
If \(y=\frac8{17}\), then
\(r=\frac{5\left(\frac8{17}\right)-4}{\frac8{17}} =\frac{\frac{40}{17}-\frac{68}{17}}{\frac8{17}} =\frac{-\frac{28}{17}}{\frac8{17}} =-\frac72.\)
If \(y=2\), then
\(r=\frac{10-4}{2}=3.\)
