Respuesta :
Answer:
The smallest possible value for the third term of the geometric progression is 1.
Step-by-step explanation:
Given : A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression.
To find : What is the smallest possible value for the third term of the geometric progression?
Solution :
The arithmetic progression is given by [tex]a,a+d,a+2d[/tex]
First term a=9
Second term - 2 is added to second term i.e. [tex]a+d+2=9+d+2=11+d[/tex]
Third term - 20 is added to the third term i.e.[tex]a+2d+20=9+2d+20=29+2d[/tex]
The geometric progression is given by [tex]a,ar,ar^2[/tex]
First term a=9
Second term -[tex]ar=11+d[/tex]
Third term [tex]ar^2=29+2d[/tex]
r is the common ratio which is second term divided by first term,
So, [tex]r=\frac{ar}{a}=\frac{11+d}{9}[/tex]
or third term divided by second term,
So, [tex]r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}[/tex]
Equating both the r,
[tex]\frac{11+d}{9}=\frac{29+2d}{11+d}[/tex]
Cross multiply,
[tex](11+d)\times (11+d)=(29+2d)(9)[/tex]
[tex]11^2+11d+11d+d^2=261+18d[/tex]
[tex]121+22d+d^2-261-18d=0[/tex]
[tex]d^2+4d-140=0[/tex]
Solving by middle term split,
[tex]d^2+14d-10d-140=0[/tex]
[tex]d(d+14)-10(d+14)=0[/tex]
[tex](d+14)(d-10)=0[/tex]
[tex]d=-14,10[/tex]
Substituting the value of d in the third term,
Third term [tex]ar^2=29+2d[/tex]
When d=-14,
Third term [tex]ar^2=29+2(-14)=29-28=1[/tex]
When d=10,
Third term [tex]ar^2=29+2(10)=29+20=49[/tex]
Therefore, The smallest possible value for the third term of the geometric progression is 1.