Respuesta :
Answer:
Step-by-step explanation:
The first four terms of geometric series is:
[tex]a,ar,ar^2,ar^3[/tex]
Since, we have given information that the third number is greater than 44 that means:
[tex]ar^2=a+44[/tex]
Above equation can be rewritten as:
[tex]a(r^2-1)=44[/tex]
Now, using:
[tex]a^2-b^2=(a+b)(a-b)[/tex]
Here, a=r,b=1
[tex]a(r+1)(r-1)=44[/tex] (1)
The sum of first four terms is:
[tex]a+ar+ar^2+ar^3=220[/tex]
[tex]a(1+r+r^2+r^3)=220[/tex]
[tex]a(1(1+r)+r^2(1+r))=220[/tex]
[tex]a((1+r)(1+r^2))=220[/tex] (2)
Divide equation (2) by (1) we get:
[tex]\frac{r^2+1}{r-1}=\frac{220}{44}[/tex]
[tex]r^2+1=5r-5[/tex]
[tex]\Rightarrow r^2-5r+6=0[/tex]
[tex]\Rightarrow r^2-3r-2r+6=0[/tex]
[tex]\Rightarrow r(r-3)-2(r-3)=0[/tex]
[tex]\Rightarrow (r-2)(r-3)=0[/tex]
[tex]\Rightarrow r=2,3[/tex]
CASE1: When r=2 in [tex]ar^2=a+44[/tex]
[tex]a(4)=a+44[/tex]
[tex]3a=44[/tex]
[tex]a=\frac{44}{3}[/tex]
CASE2:When r=3 in [tex]ar^2=a+44[/tex]
[tex]a(3)^2=a+44[/tex]
[tex]\Rightarrow 9a=a+44[/tex]
[tex]\Rightarrow 8a=44[/tex]
[tex]\Rightarrow a=\frac{44}{8}=\frac{11}{2}[/tex]
The series becomes:
From CASE1: [tex]\frac{44}{3},\frac{44\cdot 2}{3},\frac{44\cdot 2^2}{3},\frac{44\cdot 2^3}{3}[/tex]
[tex]\Rightarrow \frac{44}{3},\frac{88}{3},\frac{176}{3},\frac{352}{3}[/tex]
From CASE2: [tex]\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}[/tex]
[tex]\Rightarrow \frac{11}{2},\frac{33}{2},\frac{99}{2},\frac{297}{2}[/tex]
A geometric progression is characterized by a common ratio.
The terms of the progression are: [tex]\mathbf{T_n =5.5, 16.5, 49.5,148.5}[/tex] or [tex]\mathbf{T_n =\frac{44}{3}, \frac{88}{3}, \frac{176}{3},\frac{352}{3}}[/tex]
The given parameters are:
[tex]\mathbf{S_4 = 220}[/tex]
[tex]\mathbf{T_3 = T_1 + 44}[/tex]
The third term is represented as:
[tex]\mathbf{T_3 = ar^2}[/tex]
The first term is:
[tex]\mathbf{T_1 = a}[/tex]
So, we have:
[tex]\mathbf{ar^2 = a + 44}[/tex]
Subtract a from both sides
[tex]\mathbf{ar^2 - a = 44}[/tex]
Factor out a
[tex]\mathbf{a(r^2 - 1) = 44}[/tex]
Express as difference of two squares
[tex]\mathbf{a(r + 1)(r - 1) = 44}[/tex]
Make r + 1 the subject
[tex]\mathbf{r +1 = \frac{44}{a(r -1)}}[/tex]
Recall that:
[tex]\mathbf{S_4 = 220}[/tex]
This gives:
[tex]\mathbf{a + ar + ar^2 + ar^3 = 220}[/tex]
Factor out a
[tex]\mathbf{a(1 + r + r^2 + r^3) = 220}[/tex]
Factor out r^2
[tex]\mathbf{a(1 + r + r^2(1 + r)) = 220}[/tex]
Rewrite as:
[tex]\mathbf{a(1(1 + r) + r^2(1 + r)) = 220}[/tex]
Factor out 1 + r
[tex]\mathbf{a((1 + r^2)(1 + r)) = 220}[/tex]
Substitute [tex]\mathbf{r +1 = \frac{44}{a(r -1)}}[/tex] in [tex]\mathbf{a((1 + r^2)(1 + r)) = 220}[/tex]
[tex]\mathbf{a((1 + r^2)\times \frac{44}{a(r -1)} = 220}[/tex]
[tex]\mathbf{((1 + r^2)\times \frac{44}{(r -1)} = 220}[/tex]
Divide both sides by 44
[tex]\mathbf{(1 + r^2)\times \frac{1}{(r -1)} = 5}[/tex]
Multiply through by r - 1
[tex]\mathbf{1 + r^2 = 5r - 5}[/tex]
Collect like terms
[tex]\mathbf{r^2 - 5r + 5 +1 = 0}[/tex]
[tex]\mathbf{r^2 - 5r + 6 = 0}[/tex]
Expand
[tex]\mathbf{r^2 - 2r - 3r + 6 = 0}[/tex]
Factorize
[tex]\mathbf{r(r - 2) - 3(r - 2) = 0}[/tex]
Factor out r -2
[tex]\mathbf{(r - 3)(r - 2) = 0}[/tex]
So, we have:
[tex]\mathbf{r = 3\ or\ r = 2}[/tex]
Calculate a using: [tex]\mathbf{a(r^2 - 1) = 44}[/tex]
When r = 3
[tex]\mathbf{a(3^2 -1) = 44}[/tex]
[tex]\mathbf{a(9 -1) = 44}[/tex]
[tex]\mathbf{8a = 44}[/tex]
[tex]\mathbf{a = 5.5}[/tex]
When r = 2
[tex]\mathbf{a(2^2 -1) = 44}[/tex]
[tex]\mathbf{a(4 -1) = 44}[/tex]
[tex]\mathbf{3a = 44}[/tex]
[tex]\mathbf{a = \frac{44}{3}}[/tex]
The nth term of a GP is:
[tex]\mathbf{T_ = ar^{n-1}}[/tex]
So, the terms of the progression are:
[tex]\mathbf{T_n =5.5, 16.5, 49.5,148.5}[/tex]
or
[tex]\mathbf{T_n =\frac{44}{3}, \frac{88}{3}, \frac{176}{3},\frac{352}{3}}[/tex]
Read more about geometric progressions at:
https://brainly.com/question/18109692
