if a, b, c, d are in GP, then show that a2+b2, b2+c2, c2+d2 are in GP.
Answers
Answer:
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Step-by-step explanation:
From the question, it is given that
a,b,c,d are in G.P
So,bc=ad
b
2
=ac
c
2
=bd… We have to show that,
a
2
+b
2
,b
2
+c
2
,c
2
+d
2
are in G.P.
Then, (b
2
+c
2
)
2
=(a
2
+b
2
)(c
2
+d
2
)
Consider the LHS=
(b
2
+c
2
)
2
=b
4
+c
4
+2b
2
c
2
From the equation (ii) and equation (iii)
=a
2
c
2
+b
2
d
2
+a
2
d
2
+b
2
c
2
=c
2
(a
2
+b
2
)+d
2
(a
2
+b
2
)
=(a
2
+b
2
)(c
2
+d
2
)
Now consider the
RHS=(a
2
+b
2
)(c
2
+d
2
)
By comparing the LHS and RHS LHS=RHS
Hence it is proved that, a
2
+b
2
,b
2
+c
2
,c
2
+d
2
are in G.P.
(ii) (b−c)
2
+(c−a)
2
+(d−b)
2
=(a−d)
2
From the question, it is given that
a,b,c,d are in G.P We have to prove that,
(b−c)
2
+(c−a)
2
+(d−b)
2
=(a−d)
2
Consider the LHS=(b−c)
2
+(c−a)
2
+(d−b)
2
We know that the first, second, and third terms of G.P.
generally a, ar, ar
2
So,LHS
=(ar−ar
2
)
2
+(ar
2
−a)
2
+(ar
3
−ar)
2
=a
2
r
2
(1−r)
2
+a
2
(r
2
−1)
2
+a
2
r
2
(r
2
−1)
2
By taking out a
2
as common we get,
=a
2
[r
2
(1−r
2
−2ar)+r
4
−2r
2
+1+r
2
(r
4
−2r
2
+1)]
=a
2
[r
2
−r
4
−2ar
3
+r
4
−2r
2
+1+r
6
−2r
4
+r
2
]
=a
2
(r
6
−2r
3
+1)
Now, consider the RHS=(a−d)
2
=(a−ar
3
)
2
=a
2
(1−r
3
)
2
=a
2
(1+r
6
−2r
3
)
=a
2
(r
6
−2r
3
+1)
By comparing the LHS and RHS LHS=RHS
Hence it is proved that, (b−c)
2
+(c−a)
2
+(d−b)
2
=(a−d)
2