Determined the sum of the geometric series for a1 =-486 r= 1/3 an=-6

Question

A1=-486 r=1/3 an =-6

Answer 1

a_n = a₁(r)^n-1

-6 = -486(1/3)^n-1

1/81 = (1/3)^n-1

(1/3)⁴ = (1/3)^n-1

4 = n - 1

5 = n

S_n = a₁[(1 - rⁿ) / (1 - r)]

S₅ = -486[(1 - (1/3)⁵) / ((1 - (1/3))]

S₅ = -486[(1 - (1/243)) / (2/3]

S₅ = -486[(242/243) / (2/3)]

S₅ = -486[121/81]

S₅ = -726