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find the sum of the first six terms of a geometric progression .​

Sagot :

MrRoyal

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]

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