The fifth term of a geometric series is 405 and the sixth term is 1215. Find the sum of the first nine terms.

Question

Don't know how to find the first term please explain. 

Answer 1

r = a_n+1 / a_n

r = 1215 / 405

r = 3

a_n = a₁(r)^n-1

405 = a₁(3)⁴

405 = a₁(81)

5 = a₁

S_n = a₁(1 - rⁿ) / (1 - r)

S₉ = 5(1 - 3⁹) / (1 - 3)

S₉ = 5(1 - 19683) / (-2)

S₉ = -98410 / -2

S₉ = 49250

Answer 2

Emauel, a geometric series means we are continually multiplying by a fixed number. 1, 5, 25, 125,... is a geometric series.

 

Since 1215 = 405(3), let's work backward. 405/3 = 135

135/3 = 45

45/3 = 15

15/3 = 5

 

We also need to go the other way.

1215(3) = 3645

3645(3) = 10935

10935(3) = 32805

 

Add all 9 terms for your answer.