Respuesta :

loneguy

[tex]\text{First term,} ~a = 72\\\\\text{Common ratio,}~ r = \dfrac{-36}{72} =-\dfrac 12 \\\\\text{Sum of an infinite geometric series,}\\\\S_{\infty} = \dfrac a{1 - r}\\\\\\~~~~~= \dfrac{72}{1+ \dfrac 12}\\\\\\~~~~~=\dfrac{72 }{\tfrac 32}\\\\\\~~~~~=\dfrac{72 \times 2}3\\\\\\~~~~~=48[/tex]

fieryanswererft

Series: 72 - 36 + 18 - 9 +...

[tex]\sf S \infty \ =a_1 \ x \ \dfrac{1}{1-r}[/tex]

Identify the following

  • a1 = 72 - 36
  • r = (18-9) ÷ (72-36) = 1/4

Solve:

[tex]\rightarrow \sf S \infty \ =(72-36) \ x \ \dfrac{1}{1-\frac{1}{4} }[/tex]

[tex]\rightarrow \sf S \infty \ =48[/tex]

This series has a sum of 48.

Otras preguntas