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If a, b, c ,d be in G.P. , show that (i) (b -c)^(2) + (c - a)^(2) +(d -b)^(2) = (a - d)^(2) (ii) a^(2) + b^(2) + c^(2) , ab + bc + cd , b^(2) + c^(2) + d^(2)are in G.P. |
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Answer» Solution :(i) Let r be the common ratio of the G.P. than B = ar , ` ar^(2) and d = ar^(3)` L.H.S =`(b-c)^(2) + (c-a)^(2) + (d-b)^(2)` ` = (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)` `a^(2) [r^(2) (1 + r^(2) -2R) + (r^(4) + 1 - 2r^(2)) + r^(2) (r^(4) + 1 - 2r^(2))]` =`a^(2) [r^(2) 1 + r^(4) -2r^(3) + r^(4) + 1 - 2r^(2) +r^(6) + r^(2) - 2r^(4)]` `=a^(2)[r^(6) + 2r^(2) - 1]` ` a^(2) (r^(3) -1)^(2)` `= [a(r^(3) -1)]^(2)` = `(ar^(3) -a)^(2)` ` (d -a)^(2)` ` = (a -d)^(2)` Let r be the common ratio of the G.P. than`b = ar , c ar^(2) and d = ar^(3)` Now, ` a^(2) + b^(2) + c^(2) = a^(2) +a^(2) r^(2) + a^(2) r^(4) = a^(2) (1 + r^(2) + r^(4))` `ab + bc + CD = a^(2) r + a^(2) r^(3) + a^(2) t^(5) = a^(2)r(1 + t^(2) + r^(4))` ` b^(2) + c^(2)+d^(2) =a^(2)r^(2) + a^(2) r^(4) + a^(2) t^(6) = a^(2)r^(2)(1 + t^(2) + r^(4))` Clearly , `(ab + bc + cd)/(a^(2)+ b^(2) + c^(2))=(b^(2) + c^(2)+d^(2))/(ab + bc + cd)` Hence , ` a^(2) + b^(2) + c^(2) , ab + bc + cd , b^(2) + c^(2) + d^(2)` are in G.P. |
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