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Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]`Let b=6, with a and c satisfying (E). If alpha and beta are the roots of the quadratic equation `ax^2+bx+c=0 then sum_(n=0)^oo (1/alpha+1/beta)^n` is (A) 6 (B) 7 (C) `6/7` (D) ooA. 6B. 3C. `6/7`D. `infty` |
Answer» Correct Answer - B `because b= 6, ` with a and c satisgying ( E) `therefore a + 48 + 7 c = 0, 9a + 12 + 3c= 0, a + 6 +c=0` we get `a = 1, c -7` Given, `alpha ,beta` are the roots of `ax^(2) + bx+c=0` `therefore alpha + beta = -b/a = -6, ` `alpha beta = c/a = -7` Now, `1/alpha + 1/beta=(alpha + beta)/(alpha beta) = (-6)/(-7) = 6/7` `therefore sum _(n=0) ^(infty) (1/alpha +1/beta)^(n) = sum _(n=0)^(infty)(6/7)^(n) ` `= 1+ ( 6/7) + (6/7) ^(2) +...infty` `1/(1-6//7)=7` |
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