Chapter 12.3, Problem 79E

St. Ives The following is a well-known children’s rhyme:

As I was going to St. Ives,

I met a man with seven wives;

Every wife had seven sacks;

Every sack had seven cats;

Every cat had seven kits;

Kits, cats, sacks, and wives,

How many were going to St. Ives?

Assuming that the entire group is actually going to St. Ives, show that the answer to the question in the rhyme is a partial sum of a geometric sequence, and find the sum.

To determine

Whether the given rhyme forms a partial sum of a geometric sequence and the sum.

Answer to Problem 79E

The given rhyme forms a partial sum of a geometric sequence and the sum is 2802.

Explanation of Solution

It is given that a man has 7 wives that is the first term of the sequence is 7.

Every wife has 7 sacks.

So the total number of sack is,

$\begin{array}{c}a\times 7=7\times 7\\ =49\end{array}$

The second term of the sequence is 49.

Every sack contains 7 cats.

So total number of cat is,

$49\times 7=343$

The third term of the sequence is 343.

Every cat has 7 kits.

So total number of kits is,

$343\times 7=2401$

The sequence formed is $7,49,343,2401$.

The first term of the sequence is 7.

The ratio of the first two terms is,

$\begin{array}{c}\frac{a_2}{a_1}=\frac{49}{7}\\ =7\end{array}$

And the ratio of the next two terms is,

$\begin{array}{c}\frac{a_3}{a_2}=\frac{343}{49}\\ =7\end{array}$

Since the ratio is same, the sequence is a geometric sequence with common ratio as 7 and first term as 7.

The formula for nth partial sum of a geometric sequence is given below,

$S_n=a\frac{r^n-1}{r-1}$

Substitute 7 for a, 7 for r, and 4 for n in the above formula.

$\begin{array}{c}{S}_4=7\left(\frac{{\left(7\right)}^4-1}{7-1}\right)\\ =\frac{7}{6}\left(2401-1\right)\\ =\frac{7}{6}\left(2400\right)\\ =2800\end{array}$

The group includes a boy and a man too.

So, the total members in the group is $2800+1+1=2802$.

Thus, the given rhyme forms a partial sum of a geometric sequence and the sum is 2802.