(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.
(a) (i) ar\(^2\) = \(\frac{1}{4}\) ........(i)
ar\(^5\) = \(\frac{1}{32}\) ........(ii)
\(\frac{ar^5}{ar^2} = \frac{1}{32} \div \frac{1}{4}\)
r\(^3 = \frac{1}{32} \times \frac{4}{1}\)
r\(^3 = \frac{1}{32} \times \frac{4}{1}\)
r\(^3 = 2^{-3} = (\frac{1}{2})^3\)
r = \(\frac{1}{2}\)
Since ar\(^2\) = \(\frac{1}{4}\)
a(\(\frac{1}{2}\))\(^2\) = \(\frac{1}{4}\)
a = 1
The first term is 1 ad the common ratio is \(\frac{1}{2}\)
(ii) T\(_7\) = ar\(^6\) = 1 x (\(\frac{1}{2}\))\(^6\)
The seventh term = \(\frac{1}{64}\)
(b) x = 2 and x = -3
a(2)\(^2\) + b(2) + c = 0
4a + 2b + c = 0
4a + 2b + c = 0.....(i)
a(-3)\(^2\) + b(-3) + c = 0
9a - 3b + c = 0.......(ii)
Solving eq.(i) and (ii)
4a + 2b + c = 0
-(9a - 3b + c) = 0
-5a + 5b = 0
(x - 2)(x + 3) = 0
x\(^2\) + 3x - 2x - 6 = 0
x\(^2\) + x - 6 = 0
a = 1, b = 1, c = -6
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