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 in terms of a for the sum to infinity of the geometric series whose first term are a,c,b

honestly my soul is ripped off so hard
step plz

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1. ripped off? Goodness
Well the arith series tells us that
b=a+d, and c=a+2d
the geo series tells us that
c=ar, and b=ar^2

and we are looking for r.
ar=a+2d
ar^2=a+d
a+2d)r=a+d
r=(a+d)/(a+2d)=(1+d/a)/(1+2d/a)

so now, the sum:
Sum from n=1 to N

Sum=a(1-r^n)/(1-r)
now if n is inf, the r^n=0 because r is less than zero (see r above).
sum=a(1)/(1-(a+d)/(a+2d) )

sum= a(a+2d)/(a+2d-a-d)
sum= a(a+2d)/d

check that all, it is easy to err when doing these on a computer screen.

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2. Here is my attempt:
a,b, and c are in AS, so
b-a = c-b
c = 2b-a

a, c, b are in GS, so
c/a = b/c
c^2 = ab
(2b-a)^2 = ab
4b^2 - 4ab + a^2 = ab
4b^2 - 5ab + a^2 = 0
(b-a)(4b - a) = 0
b = a or b = a/4

but the numbers are supposed to be distinct, so can only go with
b = a/4

also c^2 = ab
= a(a/4) = a^2/4
c = ± a/2

common ratio of the GS a, c, b is
c/a
= (± a/2) / a = ± 1/2
so r = ± 1/2

sum(all terms) = a/(1-r)= a/(1 - 1/2)
= a/(1/2) = 2a
or
sum(all terms) = a/(1 + 1/2)
= a/(3/2) = (2/3)a

let's test this:
let a = 16, then b = 16/4 = 4
and c^2 = ab = 64
c = ±8
the AS is 16, 4, -8
the GS = 16, -8, 4 ---> r = -1/2
sum(all terms) = 16/(1+1/2) = 32/3 which is (2/3)a

let a = -12, b = -12/4 = -3
c^2 = ab = 36
c = ± 6
r = c/a = ±6/12 = ± 1/2

the AS is -12, -3, 6
the GS = -12, 6, -3 ---> r = -1/2
sum(all terms) = -12/(1+1/2) = -8
which is (2/3)a

notice that since c^2 = ab
a and b must be the same sign, to get a positive value of c^2

which means r = -1/2 always

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posted by Reiny

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