posted by .

There are two positive numbers that can be inserted between 3 and 9 such that the first three are in geometric progression while the last three are in arithmetic progression. Find the sum of those two numbers.

  • math -

    let the sequence be
    3 a b 9

    then b/a = 9/b and b-a = 9-b
    from the first...
    b^2 = 9a
    a = b^2/9

    from the second...
    a = 2b-9

    so b^2/9 = 2b-9
    b^2 = 18b - 81
    b^2 - 18b + 81 = 0
    (b-9)(b-9) = 0
    b = 9
    then a = 9

    sum of these two numbers is 18

  • math -

    Thanks, this helped a lot! :)

  • ERROR - math -

    My solution is wrong, I misread the question.

    should be
    b/a = a/3 ---- a^2 = 3b

    b-a = 9-b ---> a = 2b-9

    then (2b-9)^2 = 3b
    4b^2 - 36b + 81 = 3b
    4b^2 -39b + 81 = 0
    b = (39 ± √225)/8
    = 27/4 or 3

    if b=3, then a = 3, sum of those two is 6
    if b = 27/4, then a = 9/2 or 4.5 , their sum is

    however.... 3 3 3 9 does not satisfy the original condition, while

    3 , 9/2 , 27/4 , 9 does work

    so the 2 numbers inserted are 9/2 and 27/4
    or 4.5 and 6.75
    their sum is 11.25

    sorry about the previous post.

  • math -

    Ah, okay I see. Thanks for correcting it.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math (Geometric Progression)

    5 distinct positive reals form an arithmetic progression. The 1st, 2nd and 5th term form a geometric progression. If the product of these 5 numbers is 124 4/9, what is the product of the 3 terms of the geometric progression?
  2. Algebra/Number Theory

    In a sequence of four positive numbers, the first three are in geometric progression and the last three are in arithmetic progression. The first number is 12 and the last number is 452. The sum of the two middle numbers can be written …
  3. math

    The first, the third and the seventh terms of an increasing arithmetic progression are three consecutive terms of a geometric progression. In the first term of the arithmetic progression is 10 find the common difference of the arithmetic …
  4. Math

    Three numbers form a geometric progression. If 4 is subtracted from the third term, then the three numbers will form an arithmetic progression. If, after this, 1 is subtracted from the second and third terms of the progression, then …
  5. math

    The sum of the 3 terms in arithmetic sequence is 39.if these numbers is increased by 1,5 and 12 respectively the numbers would be in geometric progression. find the second term of the geometric progression?
  6. Maths

    The numbers p,10 and q are 3 consecutive terms of an arithmetic progression .the numbers p,6 and q are 3 consecutive terms of a geometric progression .by first forming two equations in p and q show that p^2-20p+36=0 Hence find the …
  7. plz sequence sir steve reiny bob damon i need u

    The three real,distint and non-zero numbers a,b,c are such that a,b,c are in arithmetic progression and a,c,b are in geometric progression. Find the numerical value of the common ratio of the geometric progression.. Hence,find an expression …
  8. math

    Three numbers form a geometric progression. If the second term is increased by 2, then the progression will become arithmetic and if, after this, the last term is increased by 9, then the progression will again become geometric. Find …
  9. algebra

    The sum of the three numbers in Arithmetic Progression is 33. If the numbers are increased by 2, 1, and 6 respectively the new numbers will be in Geometric progression. Find these numbers.
  10. math!!

    Three numbers are in an arithmetic progression; three other numbers are in a ge- ometric progression. Adding the corresponding terms of these two progressions yields 32, 26, and 32. The sum of the three terms of the arithmetic progression …

More Similar Questions