Consider three numbers x, y, and z such that x<y<z and x + y
+2=6/5. If x, y, and z are in arithmetic progression and x?.y?.
and z2 are in geometric progression, then what is the value of
Z?
Answers
Given : three numbers x, y, and z such that x<y<z and x + y+z=6/5.
x, y, and z are in arithmetic progression
x², y² and z² are in geometric progression
To Find : the Value of z
Solution:
x + y + z = 6/5
x + z = 2y
3y = 6/5
=> y = 2/5
x + z = 4/5
x² , y² and z² are in GP
=> x².z² = (y²)²
=> x².z²= ((2/5)²)²
xz = ±4/25
x + z = 4/5 , xz = ±4/25
t² - (4/5)t + 4/25= 0 , t² - (4/5)t - 4/25= 0
value of t will give value of x and z .
t² - (4/5)t + 4/25= 0
=> (t - 2/5)² = 0
=> t = 2/5
Hence x and z are 2/5
=> x , y and z are 2/5
which is not possible
as x < y < z
t² - (4/5)t - 4/25= 0
=> t² - (4/5)t - 4/25 + 8/25 = 8/25
=> (t - 2/5)² = 8/25
=> t - 2/5 = ±2√2/5
=> t = (2 ± 2√2)/5
x < z
Hence x = (2 - 2√2)/5 and z = (2 + 2√2)/5
Value of z is (2 + 2√2)/5
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