if a,b,c are in are in a continued proportion the expression { a^2 +ab+b^2/b^2+bc+c^2 } can be simplified
Answers
its obviously a^2/b^2
still
a^2 + b^2 + ab)/( b^2 + bc + c^2)
a^2/b^2 + 1+ a/b)/(1+ c/b + c^2/b^2)
a^2/b^2 +1 + a/b )/( 1+ b/a + b^2/a^2)
a^2 + b^2 + ab )a^2/( a^2 + ab + b^2)b^2
a^2/b^2
Answer:
To Prove : (a+b+c)(a−b+c)=a
2
+b
2
+c
2
Proof : a,b,c are in continued proportion.
∴
b
a
=
c
b
=k (let)
b=ck
a=bk=(ck)k =ck
2
L.H.S. =(ck
2
+ck+c)(ck
2
−ck+c)
=c
2
(k
2
+k+1)(k
2
−k+1)
=c
2
[(k
2
+1)
2
−(k)
2
]
=c
2
[k
4
+2k
2
+1−k
2
]
=c
2
[k
4
+k
2
+1]
R.H.S. =c
2
k
4
+c
2
k
2
+c
2
=c
2
[k
4
+k
2
+1]
L.H.S = R.H.S.
Step-by-step explanation:
radhe radhe