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
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.
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
I understand that this problem may seem challenging, but we can break it down into smaller steps to find the solution. Let's proceed step by step:
Step 1: Define the arithmetic progression
In an arithmetic progression, the difference between any two consecutive terms is constant. Let's say the three numbers in arithmetic progression are a, b, and c. Therefore, we can express them as:
b = a + d (Equation 1)
c = a + 2d (Equation 2)
Here, 'd' represents the common difference between the terms.
Step 2: Define the geometric progression
In a geometric progression, each term is obtained by multiplying the previous term by a constant called the common ratio (r). Since a, c, and b are in a geometric progression, we have:
c = ar (Equation 3)
Step 3: Find the common ratio (r)
Substituting Equation 2 and Equation 3 into Equation 1, we can solve for the value of r:
a + 2d = a * r
Rearranging the equation, we get:
r = (a + 2d) / a
Step 4: Simplify the common ratio expression
To simplify the common ratio expression, we need to eliminate 'd'. We can do this by substituting Equation 3 into Equation 2 and then rearranging the equation to express 'd' in terms of 'a':
c = a + 2d
ar = a + 2d
2d = ar - a
d = a(r - 1)
Substituting this expression for 'd' into the common ratio equation:
r = (a + 2d) / a
r = (a + 2(a(r - 1))) / a
r = (3ar - 2a) / a
r = 3r - 2
We can now solve this equation for 'r':
r - 3r = -2
-2r = -2
r = 1
Step 5: Find the sum to infinity of the geometric series
The sum to infinity of a geometric series can be found using the formula:
S = a / (1 - r)
Substituting our value of 'r' into the formula, we get:
S = a / (1 - 1)
S = a / 0
Since the denominator is zero, the sum to infinity of the geometric series is undefined.
Therefore, the numerical value of the common ratio is 1, but the sum to infinity of the geometric series whose terms are a, c, and b is undefined.