If 192;x;y;3 form a geometric progression, calculate the value of x and y.
3d = 3-192 = -189
d = -63
192;129;66;3
192,x,y,3,.....
t1 ,t2,t3,t4,...
if it is in GP then
(t2/t1)=(t3/t2)=(t4/t3)
then (t2/t1)=x/192.........(1)
(t3/t2)=y/x...........(2)
(t4/t3)=3/y...........(3)
Now (1)=(2)
x/192=y/x
=>192y=x^2.......(4)
Also
(2)=(3)
y/x=3/y
=>3x=y^2.........(5)
from (2)=> (y^2)/3 substitute in (4)
=>192y=((y^2)/3)^2
=>192y=(y^4)/9
=>192*9=y^3
=>1728=y^3
=>y=12 substitute in (5)
=>we get x=48
therfore x=48,y=12
then verify the series
192,48,12,3........
Hence solved.
To find the values of x and y in the geometric progression 192, x, y, 3, we can use the definition of a geometric progression. In a geometric progression, each term is obtained by multiplying the previous term by a constant ratio.
So, let's set up the following equations based on the given information:
x = 192 * r
y = x * r = (192 * r) * r = 192 * r^2
We also know that 3 = y * r, so we can substitute the value of y from the second equation into the third equation and solve for r:
3 = (192 * r^2) * r
3 = 192 * r^3
r^3 = 3/192
r = (3/192)^(1/3)
Once we have the value of r, we can substitute it back into the equations for x and y to find their respective values:
x = 192 * r
y = 192 * r^2
By plugging in the value of r into these equations, we can calculate the values of x and y in the geometric progression.