Exercise 30

A child withdraws a certain amount from his piggy bank every day. On the first day he withdrew $243, and on the last day $32. It is known that the amounts withdrawn form a geometric progression and that the total sum of the withdrawals was $665. Then, the number of withdrawals made was:

  1. 5

  2. 6

  3. 7

  4. 8

  5. 9

Try to solve it before checking the answer.

  1. 6

According to Geometric Progressions, we have:

\( a_n = a_1 r^{n – 1} \hspace{4em} \boldsymbol{(1)} \)

\( S_n = \cfrac{1 – r^n}{1 – r} a_1 \hspace{4em} \boldsymbol{(2)} \)

We know that:

\[ a_1 = 243, \]
\[ a_n = 32, \]

y

\[ S_n = 665 \]

From (1) we obtain:

\[ \begin{aligned} r^{n – 1} &= \frac{a_n}{a_1} \\[.5em] &= \frac{32}{243} \\[.5em] &= \frac{2^5}{3^5} \\[.5em] &= \left( \frac{2}{3} \right)^5 \end{aligned} \]

This is,   \( r^{n – 1} = \left( \cfrac{2}{3} \right)^5 \)

Now, let’s find the answer in a roundabout way:

Since \( r^{n – 1} = \left( \frac{2}{3} \right)^5 \), we can infer that:

\[ r = \frac{2}{3} \quad \text{ y } \quad n – 1 = 5 \]

That is:

\[ r = \frac{2}{3} \quad \text{ and } \quad n = 6 \]

Now, from (2), we prove that our inference is correct:

\[ \begin{aligned} S_n &= \frac{1 – r^n}{1 – r} a_1 \\[.5em] &= \frac{ 1 – \left( \frac{2}{3} \right)^6 }{ 1 – \frac{2}{3} } (243) \\[.5em] &= \frac{ 1 – \frac{2^6}{3^6} }{ 1 – \frac{2}{3} } \left( 3^5 \right) \\[.5em] &= \frac{ \frac{3^6 – 2^6}{3^6} }{ \frac{3 – 2}{3} } \left( 3^5 \right) \\[.5em] &= \frac{ \frac{3^6 – 2^6}{3^6} }{ \frac{1}{3} } \left( 3^5 \right) \\[.5em] &= \frac{ 3 \left( 3^6 – 2^6 \right) }{ 3^6 } \left( 3^5 \right) \\[.5em] &= \frac{ 3^6 \left( 3^6 – 2^6 \right) }{ 3^6 } \\[.5em] &= 3^6 – 2^6 \\[.5em] &= 729 – 64 \\[.5em] &= 665 \quad \text{(the right number)} \end{aligned} \]

Therefore, we conclude that, certainly, \( \boldsymbol{n = 6} \).

Remark: Finding the answer by solving equations is much more complicated.

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