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If a,b,c,d are in G.P., then the value of `(a-c)^2+(b-c)^2+(b-d)^2-(a-d)^2` is (A) `0` (B) `1` (C) `a+d` (D) `a-d` |
Answer» a=a b=ar c=a`r^2` d=a`r^3` According to question `=(a-ar^2)^2+(ar-ar^2)^2+(ar-ar^3)^2-(a-ar^3)^2` `=a^2+a^2r^4-2a^2r^2+a^2r^2+a^2r^4-2a^2r^3+a^2r^6-2a^2r^4-2a^2r^3+a^2r^2+a^2r^6-2a^2r^4-a^2-a^2r^6+2a^2r^3` =0 option A is correct. |
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