Question:
Find the sum of the first six terms of a geometric progression.
1,3,9,....
Answer:
[tex]S_6 = 364[/tex]
Step-by-step explanation:
For a geometric progression, the sum of n terms is:
[tex]S_n = \frac{a(r^n - 1)}{r - 1}[/tex]
In the given sequence:
[tex]a = 1[/tex]
[tex]r = 3/1 =3[/tex]
[tex]n = 6[/tex]
So:
[tex]S_n = \frac{a(r^n - 1)}{r - 1}[/tex]
[tex]S_6 = \frac{1 * (3^6 - 1)}{3 - 1}[/tex]
[tex]S_6 = \frac{3^6 - 1}{2}[/tex]
[tex]S_6 = \frac{728}{2}[/tex]
[tex]S_6 = 364[/tex]