Question

PLEASE HELP 20 POINTS!!

Answer 1

`\text{First term,} ~a = 72\\\\\text{Common ratio,}~ r = (-36)/(72) =-\frac 12 \\\\\text{Sum of an infinite geometric series,}\\\\S_(\infty) = \frac a{1 - r}\\\\\\~~~~~= (72)/(1+ \frac 12)\\\\\\~~~~~=(72 )/(\tfrac 32)\\\\\\~~~~~=\frac{72 * 2}3\\\\\\~~~~~=48`

Answer 2

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

`\sf S \infty \ =a_1 \ x \ (1)/(1-r)`

Identify the following

• a1 = 72 - 36

• r = (18-9) ÷ (72-36) = 1/4

Solve:

`\rightarrow \sf S \infty \ =(72-36) \ x \ (1)/(1-(1)/(4) )`

`\rightarrow \sf S \infty \ =48`

This series has a sum of 48.