answer:First, we write the equation of the line using the point-slope form: ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point through which the line passes. Given ( m = 3 ) and ( (x_1, y_1) = (5, 3) ), the equation becomes: [ y - 3 = 3(x - 5) ] [ y - 3 = 3x - 15 ] [ y = 3x - 12 ] To find the x-intercept (( y = 0 )): [ 0 = 3x - 12 ] [ 3x = 12 ] [ x = 4 ] To find the y-intercept (( x = 0 )): [ y = 3(0) - 12 ] [ y = -12 ] The sum of the x-intercept and the y-intercept: [ 4 + (-12) = -8 ] Conclusion: The sum of the x-intercept and the y-intercept is (boxed{-8}).

answer:(1) Since the sequence {a_n} satisfies a_{n+1}^2=a_na_{n+2}, it follows that frac{a_{n+2}}{a_{n+1}}= frac{a_{n+1}}{a_n}. Therefore, the sequence {a_n} is a geometric sequence with the first term frac{1}{3} and let the common ratio be q. From a_1= frac{1}{3} and a_4= frac{1}{81}, we get: frac{1}{81}= frac{1}{3}times q^3, solving this gives q= frac{1}{3}. Therefore, a_n= frac{1}{3}times( frac{1}{3})^{n-1}=( frac{1}{3})^{n}. (2) f(a_n)=log_3( frac{1}{3})^{n}=-n. Therefore, b_n=f(a_1)+f(a_2)+ldots+f(a_n)=-1-2-ldots-n=- frac{n(n+1)}{2}, and frac{1}{b_n}=-2( frac{1}{n}- frac{1}{n+1}). T_n= frac{1}{b_1}+ frac{1}{b_2}+ldots+ frac{1}{b_n}=-2left[(1- frac{1}{2})+( frac{1}{2}- frac{1}{3})+ldots+( frac{1}{n}- frac{1}{n+1})right]=-2(1- frac{1}{n+1})= frac{-2n}{n+1}, Therefore, T_{2017}= frac{-2017}{1009}. Thus, the final answers are: (1) a_n= boxed{( frac{1}{3})^{n}} (2) T_{2017}= boxed{frac{-2017}{1009}}

answer:For a number to be divisible by 11, the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions must be a multiple of 11 (including zero). The number must also have two even and two odd digits to meet the condition given. Let's consider the smallest number scenario: - Use smaller digits to minimize the number. - The thousands digit should preferably be 1 to keep the number minimal. - Balance the sums of digits in odd and even positions to either be equal or differ by multiples of 11. Possible digits to consider could be: - Thousands place: 1 (odd) - Hundreds place: 0 (even) - Other digits to balance: can be 6 (even) and 5 (odd), ensuring the tens and units places balances the requirement. Let's check the sums: - Odd position digits: 1 (thousands) + 5 (units) = 6 - Even position digits: 0 (hundreds) + 6 (tens) = 6 The difference is (6 - 6 = 0), which is divisible by 11. Thus, the number 1065 meets the criteria: - It is a four-digit number. - It has two odd digits (1 and 5) and two even digits (0 and 6). - The alternating sum of its digits equals 0, which is divisible by 11. Hence, the smallest number that meets all these conditions is boxed{1065}.

answer:Since a_{n+2}=a_{n+1}-a_n, we have a_{n+3}=a_{n+2}-a_{n+1}=a_{n+1}-a_n-a_{n+1}=-a_n, which means a_{n+6}=-a_{n+3}=a_n, indicating that the sequence {a_n} has a period of 6. Therefore, a_{2014}=a_{335times6+4}=a_4, Given a_1=3, a_2=6, and a_{n+2}=a_{n+1}-a_n, we find a_3=a_2-a_1=6-3=3, and a_4=a_3-a_2=3-6=-3. Thus, a_{2014}=a_4=-3. Hence, the correct choice is boxed{text{B}}. From the condition a_{n+2}=a_{n+1}-a_n, we deduce a_{n+6}=a_n, thereby determining the sequence is periodic. Utilizing the periodicity of the sequence allows for the solution. This problem primarily examines the calculation of sequence terms, and identifying the sequence as periodic is key to solving the problem.