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Three non-zero real numbers form a AP and the squares of these numbers taken in same order form a GP. If the possible common ratios are (2pm sqrt(k)) where k in N, then the value of [k)/(8)-(8)/(k)) is (where [] denites the greatestinteger function). |
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Answer» SOLUTION :Let number of AP are `(a-d),a,(a+d)`. According to the question, `(a-d)^(2),a^(2),(a+d)^(2)` are in GP. `:.(a^(2))^(2)=(a-d)^(2)(a+d)^(2)` `implies a^(4)=(a^(2)-d^(2))^(2)` `implies a^(4)=a^(4)+d^(4)-2a^(2)d^(2)` `implies a^(2)(a^(2)-2D^(2))=0` `implies ane0," So "a^(2)=2d^(2)` `implies a=pm sqrt(2d)"" "......(i)"` Let common ratio of GP is r. `:.r^(2)=((a+d)^(2))/((a-d)^(2))` `implies r^(2)=(a^(2)+d^(2)+2ad)/(a^(2)+d^(2)-2ad)` `implies r^(2)=(2d^(2)+d^(2)+2sqrt(d^(2)))/(2d^(2)+d^(2)-2sqrt(d^(2)))` `"" [" from EQ. (i) for"a= sqrt(2)d]` `implies r^(2)=((3+2sqrt(2))d^(2))/((3-2sqrt(2))d^(2))` `implies r^(2)=((3+2sqrt(2))(3+2sqrt(2)))/(9-8)` `implies r^(2)=(3+2sqrt(2))^(2)` `implies r^(2)=(3+sqrt(8))^(2)` `:. r=pm(3+sqrt(8))` `impliesr=3+sqrt(8) "" [:." ris positive "]` Similarly, for `a=-sqrt(2)d,` we get `r=pm(3-sqrt(8))` `implies r=(3-sqrt(8)) "" [:. " r is positive "]` COMPARE r with `3pmsqrt(K)`, we get `k=8` `[(k)/(8)-(8)/(k)]=[(8)/(8)-(8)/(8)]` `= [1-1]=[0]=0`. |
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