In the sequence of numbers an, bn is a proportional sequence in which all terms are positive

Updated on educate 2024-04-13
10 answers
  1. Anonymous users2024-02-07

    Constant. So it's proportional.

    2.It's too hard to be able to do anything.

  2. Anonymous users2024-02-06

    Let an=a1*q (n-1) bn=b1*q'(n-1)

    Then bn an=(b1 a1)*(q'/q)^(n-1)

    So the first term is q'q, the common ratio is q'A proportional series of q.

  3. Anonymous users2024-02-05

    Let the first term of the proportional series be a1 and the common ratio be q

    A proportional series where each item is positive, a1>0 q>0

    an/a(n-1)=q

    an a(n-1) = q, which is a fixed value.

    A series is a proportional series with a1 as the first term and q as the common ratio.

  4. Anonymous users2024-02-04

    and a1=b1=1,a2+b3=6,a3+b2=5,1) to find the first n terms and sn of the series

  5. Anonymous users2024-02-03

    If the ratio of the public to the closed celery is q, then q > 0

    a3=a2+4

    a1q^2=a1q+4

    a1 = 2 substitution into the state Jane, finishing, getting.

    q^2-q-2=0

    q+1)(q-2)=0

    q=-1 (rounded) or q=2

    sn=a1(q n -1) (q-1)=2 (2 n -1) (2-1)=2 (n+1) -2

  6. Anonymous users2024-02-02

    Well, I just got the question wrong.

    a1+a2=2*(1/a1+1/a2) →a1(1+q)=2*(1/a1)(1+1/q) (1)

    a3+a4=32*(1/a3+1/a4) →a1*q*q*(1+q)=32*(1/a1)*(1/q^2)(1+1/q) (2)

    Because an is an integer-proportional series, (2) (1) q*q=16 q2 can be obtained q=2; q = -2 (rounding).

    Substituting equation (1) yields a1=1; a1=-1 (rounded).

    Therefore an=2 (n-1).

  7. Anonymous users2024-02-01

    a3=a1*q 2=e (b2)=e 18a6=a1*q 5=e (b6)=e 12, then a6 a3=q 3=e 12 e 18=e (-6) gives q=e (-2), a1=e 22

    The general formula for proportional series is e (24-2n).

    The sequence satisfies bn=ln(an).

    Then the general formula for the series is (24-2n).

    When n=12, bn=0

    When n>=12, bn<0

    So when the first n terms and sn are taken as the maximum, n = 12

    Then sn(n=12)=(b1+b12)*12 2=132

  8. Anonymous users2024-01-31

    b3=lna3=ln(a1*q^2)=lna1+2lnq=18 (1)

    b6=lna6=ln(a1*q^5)=lna1+5lnq=12 (2)

    2)-(1) 3lnq=-6

    So lnq=-2

    then LNA1=22

    Therefore, bn=ln an=ln[a1*q (n-1)]=lna1+(n-1)lnq

    22-2(n-1)

    24-2n let bn=24-2n 0, and the solution is n 12

    So sn max = s12

    b1+b12)*12/6

  9. Anonymous users2024-01-30

    This is because all the items in the proportional sequence are positive numbers that are not equal to 1.

    then an=a1xq (n-1) a1>0 q>0

    Then bn=in, AN=LNA1+(n-1)1nq, 1nq is constant, and bn is the equal difference series, d=1nq, b1=LNA1

    again b3=18, b6=12 b6-b3=3d=-6 d=lnq=-2 so q=e -2

    b3=lna3=lna1+2x(-2) to get a1=24 b1=lna1=ln24

    bn=ln24-2（n-1）

    The summation formula of the equal difference series is substituted.

    If the maximum value is because d<0 so bn is a decreasing series Let bn=ln24-2(n-1)=0 find n and then use the summation formula to calculate the maximum value Come on with yourself, the higher number series is the foundation to learn well, and it is beneficial to do more problems.

  10. Anonymous users2024-01-29

    The common ratio of the proportional series is denoted as q, and the sum of the first n terms of the series is denoted as snbn=lnan=ln[a1q (n-1)]=lna1+(n-1)lnq

    lna1+(3-1)lnq=18

    lna1+(6-1)lnq=12

    Solve the system of equations about LNA1 and LNQ.

    lna1=22，lnq=-2

    So bn=22+(n-1) (2).

    sn=n 22+n(n-1) 2 (-2)=-n +23n, because 23 2-11=12-23 2

    So s11=s12=-11 +23 11=132 is the maximum value of sn

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