(a) Given that (7 -2x), 9, (5x + 17) are consecutive terms of a Geometric Progression (G. P) with common ratio, r>0, find the values of x.
(b) Two positive numbers are in the ratio 3:4. The sum of thrice the first number and twice the second is 68. Find the smaller number.
(a) r =\(\frac{9}{7-2x}\)
r = \(\frac{5x+1}{7-2x}\)
\(\frac{9}{7-2x}\) = \(\frac{5x+1}{7-2x}\)
9x9=(7-2x) (5x +17)
81 = 35x + 119 - 10x\(^2\) - 34x
10x\(^2\) - x - 38
(x -2)(10x+19) = 0
x = 2, x = \(\frac{19}{10}\)
x = 2
(b) Let the integers be x and y
x:y= 3:4
\(\frac{x}{y}\) = \(\frac{3}{4}\)
x = \(\frac{3y}{4}\)
3x + 2y=68
3( \(\frac{3}{4}\)y)+ 2y = 68
\(\frac{9}{4}\)y + 2y = 68
9y + 8y = 272
17y = 272
y= 16
x = \(\frac{3}{4}\) * 16 →12
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