# geometric progression word problem,

• May 28th 2009, 10:42 AM
Tweety
geometric progression word problem,
Quote:

A liquid is kept in a barrel. At the start of a year the barrel is filled with 160 litres of the liquid. Due to evaporation, at the end of every year the amount of liquid in the barrel is reduced by 15% of its volume at the start of the year.
Quote:

(a) Calculate the amount of liquid in the barrel at the end of the first year.
Liquid at the start of the year = 160
end of 1st year = 160x0.85 = 136ml

Quote:

(b) Show that the amount of liquid in the barrel at the end of ten years is approximately 31.5 litres.
Liquid at end of year 10=160×(0.85)^10=31.499 … =31.5 litres

Quote:

At the start of each year a new barrel is filled with 160 litres of liquid so that, at the end of 20 years, there are 20 barrels containing liquid.

(c) Calculate the total amount of liquid, to the nearest litre, in the barrels at the end of 20 years.
need help with part 'c'

thanks.
• May 28th 2009, 10:53 AM
VonNemo19
The first barrel will have 85% of it's total liquid after the first year, 85% of what remains the second year, 85% of what's left after the second year, and etc... Do you recognize the pattern?
• May 28th 2009, 10:59 AM
Tweety
Quote:

Originally Posted by VonNemo19
The first barrel will have 85% of it's total liquid after the first year, 85% of what remains the second year, 85% of what's left after the second year, and etc... Do you recognize the pattern?

How do I work out the first term, 'a' ?
I know I have to use this formula $s_{n} = \frac{a(1- r^{n})}{1-r}$ to work out the total amount of liquid at the end of 20 years
• May 28th 2009, 11:38 AM
Soroban
Hello, Twetty!

Quote:

A liquid is kept in a barrel.
At the start of a year the barrel is filled with 160 litres of the liquid.
Due to evaporation, at the end of every year the amount of liquid in the barrel
is reduced by 15% of its volume at the start of the year.

At the start of each year a new barrel is filled with 160 litres of liquid so that,
at the end of 20 years, there are 20 barrels containing liquid.

(c) Calculate the total amount of liquid, to the nearest litre, in the barrels
at the end of 20 years.

At the end of 20 years . . .

The first barrel has had 20 years of evaporation.
. . It contains: . $160(0.85)^{20}$ liters.

The second barrel has had 19 years of evaporation.
. . It contains: . $160(0.85)^{19}$ liters.

The third barrel has had 18 years of evaporation.
. . It contains: . $160(0.85)^{18}$ liters.

. . . . . . . . . $\vdots$

The 18th barrel has had 3 years of evaporation.
. . It contains: . $160(0.85)^3$ liters.

The 19th barrel has had 2 years of evaporation.
. . It contains: . $160(0.85)^2$ liters.

The 20th barrel has had 1 year of evaporation.
. . It contains: . $160(0.85)$ liters.

Hence, the total amount of liquid at the end of 20 years is:

. . $S \;=\;160(0.85) + 160(0.85)^2 + 160(0.85)^3 + \hdots + 160(0.85)^{19} + 160(0.85)^{20}$

. . $S \;=\;160(0.85)\underbrace{\bigg[1 + 0.85 + 0.85^2 + 0.85^3 + \hdots + 0.85^{19}\bigg]}_{\text{geometric sries}}$

The geometric series has $a = 1,\;r = 0.85,\;n = 20$
. . Its sum is: . $\frac{1-0.85^{20}}{1-0.85} \;=\;6.408269793$

Therefore: . $S \;=\;160(0.85)(6.408269793) \;=\;871.5246918 \;\approx\;872$ liters.