If a b c d are in GP then show that abc-a2 , acd-b2 , bd2-ac2 are also in GP.
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3
Answer:
Taking LHS
(a
2
+b
2
+c
2
)(b
2
+c
2
+d
2
)
putting values b=ar,c=ar
2
,d=ar
3
(a
2
+(ar)
2
+(ar
2
)
2
)((ar)
2
+(ar
2
)
2
+(ar
3
)
2
)
=a
2
(1+r
2
+r
4
)a
2
(r
2
+r
4
+r
6
)
=a
4
r
2
(1+r
2
+r
4
)
2
RHS=(ab+bc+cd)
2
=(a×ar+ar×ar
2
+ar
2
×ar
3
)
2
=(a
2
r+a
2
r
3
+a
2
r
5
)
2
=ar
2
(1+r
2
+r
4
)
2
=LHS.
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