Respuesta :
Given:
The sum of the first three terms = 12
The sum of the first six terms = (−84).
To find:
The third term of a geometric progression.
Solution:
The sum of first n term of a geometric progression is:
[tex]S_n=\dfrac{a(r^n-1)}{r-1}[/tex]
Where, a is the first term and r is the common ratio.
The sum of the first three terms is equal to 12, and the sum of the first six terms is equal to (−84).
[tex]\dfrac{a(r^3-1)}{r-1}=12[/tex] ...(i)
[tex]\dfrac{a(r^6-1)}{r-1}=-84[/tex] ...(ii)
Divide (ii) by (i), we get
[tex]\dfrac{r^6-1}{r^3-1}=\dfrac{-84}{12}[/tex]
[tex]\dfrac{(r^3-1)(r^3+1)}{r^3-1}=-7[/tex]
[tex]r^3+1=-7[/tex]
[tex]r^3=-7-1[/tex]
[tex]r^3=-8[/tex]
Taking cube root on both sides, we get
[tex]r=-2[/tex]
Putting [tex]r=-2[/tex] in (i), we get
[tex]\dfrac{a((-2)^3-1)}{(-2)-1}=12[/tex]
[tex]\dfrac{a(-8-1)}{-3}=12[/tex]
[tex]\dfrac{-9a}{-3}=12[/tex]
[tex]3a=12[/tex]
Divide both sides by 3.
[tex]a=4[/tex]
The nth term of a geometric progression is:
[tex]a_n=ar^{n-1}[/tex]
Where, a is the first term and r is the common ratio.
Putting [tex]n=3,a=4,r=-2[/tex] in the above formula, we get
[tex]a_3=4(-2)^{3-1}[/tex]
[tex]a_3=4(-2)^{2}[/tex]
[tex]a_3=4(4)[/tex]
[tex]a_3=16[/tex]
Therefore, the third term of the geometric progression is 16.
