CIE/GCE/AS/Math/P1/19/Nov/13/Q#9
Question
The first, second and third terms of a geometric progression are 3𝑘, 5𝑘 − 6 and 6𝑘 − 4
respectively.
(i) Show that 𝑘 satisfies the equation 7k2 − 48k + 36 = 0.
(i) Find, showing all necessary working, the exact values of the common
ratio corresponding to each of the possible values of k.
(ii) One of these ratios gives a progression which is convergent. Find the
sum to infinity.
Solution
i.
From the given information, we can collect following information about this Geometric
Progression (G.P).
𝑎1 = 3𝑘
𝑎2 = 5𝑘 − 6
𝑎3 = 6𝑘 − 4
Expression for Common Ratio (𝑟) in a Geometric Progression (G.P) is;
𝑎𝑛
𝑟=
𝑎𝑛−1
Hence;
𝑎𝑛
𝑟=
𝑎𝑛−1
, CIE/GCE/AS/Math/P1/19/Nov/13/Q#9
𝑎3 𝑎3
𝑟= =
𝑎3−1 𝑎2
𝑎2 𝑎2
𝑟= =
𝑎2−1 𝑎1
𝑎3 𝑎2
=
𝑎2 𝑎1
6𝑘 − 4 5𝑘 − 6
=
5𝑘 − 6 3𝑘
(6𝑘 − 4)3𝑘 = (5𝑘 − 6)2
(5𝑘)2 − 2(5𝑘)(6) + (6)2 = 18𝑘2 − 12𝑘
25𝑘2 − 60𝑘 + 36 = 18𝑘2 − 12𝑘
7𝑘2 − 48𝑘 + 36 = 0
ii.
We are required to solve the following equation obtained in (i);
7𝑘2 − 48𝑘 + 36 = 0
Question
The first, second and third terms of a geometric progression are 3𝑘, 5𝑘 − 6 and 6𝑘 − 4
respectively.
(i) Show that 𝑘 satisfies the equation 7k2 − 48k + 36 = 0.
(i) Find, showing all necessary working, the exact values of the common
ratio corresponding to each of the possible values of k.
(ii) One of these ratios gives a progression which is convergent. Find the
sum to infinity.
Solution
i.
From the given information, we can collect following information about this Geometric
Progression (G.P).
𝑎1 = 3𝑘
𝑎2 = 5𝑘 − 6
𝑎3 = 6𝑘 − 4
Expression for Common Ratio (𝑟) in a Geometric Progression (G.P) is;
𝑎𝑛
𝑟=
𝑎𝑛−1
Hence;
𝑎𝑛
𝑟=
𝑎𝑛−1
, CIE/GCE/AS/Math/P1/19/Nov/13/Q#9
𝑎3 𝑎3
𝑟= =
𝑎3−1 𝑎2
𝑎2 𝑎2
𝑟= =
𝑎2−1 𝑎1
𝑎3 𝑎2
=
𝑎2 𝑎1
6𝑘 − 4 5𝑘 − 6
=
5𝑘 − 6 3𝑘
(6𝑘 − 4)3𝑘 = (5𝑘 − 6)2
(5𝑘)2 − 2(5𝑘)(6) + (6)2 = 18𝑘2 − 12𝑘
25𝑘2 − 60𝑘 + 36 = 18𝑘2 − 12𝑘
7𝑘2 − 48𝑘 + 36 = 0
ii.
We are required to solve the following equation obtained in (i);
7𝑘2 − 48𝑘 + 36 = 0
