Question

help plz...

Answer 1

a,b,c in geometrical progression..
a^2+b^2+c^2=S^2
a+b+c=qS

show that 1/3<q^2<3

Answer 2

@ganeshie8

Answer 3

a,b,c in geometrical progression..
does that mean
a
b=ar
c=ar^2
?

Answer 4

yes

Answer 5

yes @ikram002p

Answer 6

ok then
\( \large a^2+a^2r^2+a^2r^4=s^2\)
\( \large a^2(1+ r^2+ r^4)=s^2\)


\(\large a+b+c=qs\)
\(\large a+ar+ar^2=qs\)
\(\large a(1+r+r^2)=qs\)
\(\large a^2(1+r+r^2)^2 =q^2s^2\)

thus
\(\large q^2=\dfrac { a^2(1+r+r^2)^2 }{s^2} \)
\(\large q^2=\dfrac { a^2(1+r+r^2)^2 }{a^2(1+ r^2+ r^4) } \)

\(\large q^2=\dfrac { (1+r+r^2)^2 }{(1+ r^2+ r^4) } \)

Answer 7

mmm lets see how to go far from this

Answer 8

we know
r\(\neq 0\)
r\(\neq 1\)
so
\(0<r<1\)
or
\(1<r\)

Answer 9

mmm idk how to continue xD
since i got

\(q^2>3\) for \(r>1\) xD

Answer 10

ok ill go through this part

\(0<r<1\)
\(0<r^2<1\)
\(0<r^4<1\)
but
\(0<r^4<r^2<r <1\)
mmm

Answer 11

have no clue what the rest would be , sry