Find the sum of the following geometric sequences. d) 3 terms a= 16 r=3/4

Mathematics

1 answer:

kodGreya [7K]2 years ago

7 0

Answer:

S₃ = 37

Step-by-step explanation:

given a₁ = 16 and r = $\frac{3}{4}$ , then

a₂ = a₁ × $\frac{3}{4}$ = 16 × $\frac{3}{4}$ = 12

a₃ = a₂ × $\frac{3}{4}$ = 12 × $\frac{3}{4}$ = 9

then

S₃ = a₁ + a₂ + a₃ = 16 + 12 + 9 = 37

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