(a) The third and sixth terms of a Geometric Progression (G.P) are and \(\frac{1}{4}\) and \(\frac{1}{32}\) respectively. Find: (i) the first term and the c...
Question 1 Report
(a) The third and sixth terms of a Geometric Progression (G.P) are and \(\frac{1}{4}\) and \(\frac{1}{32}\) respectively.
Find:
(i) the first term and the common ratio;
(ii) the seventh term.
(b) Given that 2 and -3 are the roots of the equation ax\(^2\) ± bx + c = 0, find the values of a, b and c.
Answer Details
(a) Geometric Progression. \(T_3=ar^{2}=\dfrac{1}{4}\) and \(T_6=ar^{5}=\dfrac{1}{32}\).
Divide: \(\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{1/32}{1/4}=\dfrac{1}{8}\Rightarrow r=\dfrac{1}{2}\).
Then \(a\left(\tfrac{1}{2}\right)^{2}=\tfrac{1}{4}\Rightarrow \tfrac{a}{4}=\tfrac{1}{4}\Rightarrow a=1\).
(i) First term \(a=1\), common ratio \(r=\dfrac{1}{2}\).
(ii) \(T_7=ar^{6}=1\cdot\left(\dfrac{1}{2}\right)^{6}=\dfrac{1}{64}\).
(b) If \(2\) and \(-3\) are roots of \(ax^{2}+bx+c=0\):
Sum of roots \(=-\dfrac{b}{a}=2+(-3)=-1\); product \(=\dfrac{c}{a}=2\times(-3)=-6\).
Taking \(a=1\): \(b=1,\ c=-6\). So \(\mathbf{a=1,\ b=1,\ c=-6}\) (equation \(x^{2}+x-6=0\)).
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