Question

the sum of four numbers in G.p is 60 and the arithmetic mean of first and the last is 18. the numbers are

Answer 1

[tex]a_1=a\\a_2=a_1r=ar\\a_3=a_2r=ar^2\\a_4=a_3r=ar^3\\S=a+ar+ar^2+ar^3=a(1+r+r^2+r^3)\\\frac{a_1+a_4}{2}=18\\\\\frac{a+ar^3}{2}=18\\\\\frac{a(1+r^3)}{2}=18\\\\r^3=\frac{36}{a}-1\\\\r=\sqrt[3]{\frac{36}{a}-1}\ [/tex]
The sum and arithmetic mean are integer, so each number of GP must be positive integer. For a=1 or 2 or 3 the common ratio is irrational, for a=4 the ratio is:
$r=\sqrt[3]{\frac{36}{4}-1}=\sqrt[3]8=2$

Checking:
S = 4 + 4×2 + 4×4 + 4×8 = 4 + 8 + 16 + 32 = 60
m = (a₁ + a₄) / 2 = (4 + 32) / 2 = 18
It's OK.
For a = 6,  9,  12, the ratio is irrational too. For a = 18, r = 1, but then
m = (18+18) / 2 = 18,  S = 18 + 18 + 18 + 18  = 72  (false)
For a = 36,  r = 0, but r must be ≠ 0


Answer: 4,  8,  16,  32

Answer 2

Answer:

Answer for the question is given in the picture

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