Question

help in qn 11

Answer 1

any help would be appreciated

Answer 2

im srry idk

Answer 3

?

Answer 4

its tricky

Answer 5

where?

Answer 6

a, b, and c are in GP so
\[b^2=ac\]
we have
\[ \large{a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}}\]
We can write this as
\[b^2= (a^{\frac{1}{x}})^x\times (c^{\frac{1}{z}})^z \]
from the relation given we can write this as
\[b^2=(b^{\frac{1}{y}})^x\times(b^{\frac{1}{y}})^z \]
so we get
\[b^2=b^{\frac{x}{y}+\frac{z}{y}}\]
we get now
\[2y=x+z\]
hence x, y, and z are in AP

Answer 7

wow thx

Answer 8

tht was a nice qn wasnt it?

Answer 9

welcome :), yeah :)