(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 ...

(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.

Answer Details

 (a) r =97−2x $\frac{\mathrm{<br>}}{}$

r = 5x+17−2x $\frac{\mathrm{<br>}}{}$

97−2x $\frac{\mathrm{<br>}}{}$ = 5x+17−2x $\frac{\mathrm{<br>}}{}$

9x9=(7-2x) (5x +17)

 81 = 35x + 119 - 10x2 ${}^{\mathrm{<br>}}$ - 34x

10x2 ${}^{\mathrm{<br>}}$ - x - 38

(x -2)(10x+19) = 0

x = 2,  x = 1910 $\frac{\mathrm{<br>}}{}$

x = 2

(b) Let the integers be x and y

x:y= 3:4

xy $\frac{\mathrm{<br>}}{}$ = 34 $\frac{3}{4}$

x = 3y4 $\frac{\mathrm{<br>}}{}$

3x + 2y=68

3( 34 $\frac{\mathrm{<br>}}{}$y)+ 2y = 68

 94 $\frac{\mathrm{<br>}}{}$y + 2y = 68

9y + 8y = 272

17y = 272

y= 16

x = 34 $\frac{\mathrm{<br>}}{}$ * 16 →12