Since the soup kitchen starts with 16 gallons and there are 128 ounces in 1 gallong, then the first term of the sequence is \(\displaystyle{a}{1}={16}{\left({128}\right)}={2048}.\)

If a quarter of the soup is served and the rest is saved for the next day, then three-fourths of the soup is left after each day. The next terms in the sequence can then be found by multiplying each term by 3/4 to get the next term. The next four terms of the sequence are then:

\(\displaystyle{a}{2}={\frac{{{3}}}{{{4}}}}{a}{1}={\frac{{{3}}}{{{4}}}}{\left({2048}\right)}={1536}\)

\(\displaystyle{a}{3}={\frac{{{3}}}{{{4}}}}{a}{2}={\frac{{{3}}}{{{4}}}}{\left({1536}\right)}={1152}\)

\(\displaystyle{a}{4}={\frac{{{3}}}{{{4}}}}{a}{3}={\frac{{{3}}}{{{4}}}}{\left({1152}\right)}={864}\)

\(\displaystyle{a}{5}={\frac{{{3}}}{{{4}}}}{a}{4}={\frac{{{3}}}{{{4}}}}{\left({864}\right)}={648}\)

The first five terms of the sequence are then 2048,1536,1152,864,648.

Since the terms have a common ratio of \(\displaystyle{r}={\frac{{{3}}}{{{4}}}}\), then the sequence is geometric. The formula for the nnth term of a geometric sequence is \(\displaystyle{a}{n}={a}{1}{r}^{{n}}-{1}\). Since a1=2048 and r=34, then the equation that represents the nnth term of the sequence is \(\displaystyle{a}{n}={2048}{\left({\frac{{{3}}}{{{4}}}}\right)}^{{{n}}}−{1}\)

For the soup to be all gone, you would need an=0 for some value of nn. Multiplying a number by \(\displaystyle{\frac{{{3}}}{{{4}}}}\) repeatedly will give a smaller and smaller number each time you multiply. The number will never equal 0 though, it will just get very close to 0. Therefore, the soup will never be gone.

If a quarter of the soup is served and the rest is saved for the next day, then three-fourths of the soup is left after each day. The next terms in the sequence can then be found by multiplying each term by 3/4 to get the next term. The next four terms of the sequence are then:

\(\displaystyle{a}{2}={\frac{{{3}}}{{{4}}}}{a}{1}={\frac{{{3}}}{{{4}}}}{\left({2048}\right)}={1536}\)

\(\displaystyle{a}{3}={\frac{{{3}}}{{{4}}}}{a}{2}={\frac{{{3}}}{{{4}}}}{\left({1536}\right)}={1152}\)

\(\displaystyle{a}{4}={\frac{{{3}}}{{{4}}}}{a}{3}={\frac{{{3}}}{{{4}}}}{\left({1152}\right)}={864}\)

\(\displaystyle{a}{5}={\frac{{{3}}}{{{4}}}}{a}{4}={\frac{{{3}}}{{{4}}}}{\left({864}\right)}={648}\)

The first five terms of the sequence are then 2048,1536,1152,864,648.

Since the terms have a common ratio of \(\displaystyle{r}={\frac{{{3}}}{{{4}}}}\), then the sequence is geometric. The formula for the nnth term of a geometric sequence is \(\displaystyle{a}{n}={a}{1}{r}^{{n}}-{1}\). Since a1=2048 and r=34, then the equation that represents the nnth term of the sequence is \(\displaystyle{a}{n}={2048}{\left({\frac{{{3}}}{{{4}}}}\right)}^{{{n}}}−{1}\)

For the soup to be all gone, you would need an=0 for some value of nn. Multiplying a number by \(\displaystyle{\frac{{{3}}}{{{4}}}}\) repeatedly will give a smaller and smaller number each time you multiply. The number will never equal 0 though, it will just get very close to 0. Therefore, the soup will never be gone.