A recipe uses 105 millilitres of sauce on day 1 and 85 millilitres on day 2. From day 3 onward, the amount used each day is the average of the amounts used on the two previous days. What is the integer part of the amount used on day 19?

Step-by-step solution

1. 1

   Work with the differences, not the terms. Subtracting $a_{n-1}$ from the recurrence gives

   $a_n-a_{n-1}=\frac{a_{n-1}+a_{n-2}}{2}-a_{n-1}=-\frac{a_{n-1}-a_{n-2}}{2}$

   So each gap is exactly $-\tfrac12$ times the previous gap. Turning the two-term recurrence into a one-term geometric one is what makes term $19$ reachable without listing everything.

2. 2

   Write the gaps explicitly. The first gap is $a_2-a_1=-20$, so

   $a_{n}-a_{n-1}=-20\left(-\frac12\right)^{n-2}$

   The alternating sign is why the terms zig-zag ($105,\ 85,\ 95,\ 90,\ 92.5,\dots$) instead of marching one way.

3. 3

   Sum the geometric series to get the limit. Adding all the gaps from $a_1$ onward:

   $L=105+\sum_{k=0}^{\infty}(-20)\left(-\frac12\right)^{k}=105-\frac{20}{1+\frac12}=105-\frac{40}{3}=\frac{275}{3}$

   Equivalently $L=\dfrac{a_1+2a_2}{3}$: the weighted average with the newer term counted twice, because the invariant $a_{n-1}+2a_n$ is unchanged by the recurrence. It is not $2a_2-a_1$.

4. 4

   Build the closed form. Since the deviation from $L$ shrinks by the factor $-\tfrac12$ each step,

   $a_n=L+(a_1-L)\left(-\frac12\right)^{n-1}=\frac{275}{3}+\frac{40}{3}\left(-\frac12\right)^{n-1}$

   Check it at $n=2$: $\tfrac{275}{3}-\tfrac{20}{3}=85$. Correct.

5. 5

   Evaluate at $n=19$. The exponent $18$ is even, so the correction is positive:

   $a_{19}=\frac{275}{3}+\frac{40}{3}\cdot\frac{1}{2^{18}}=\frac{3003735}{32768}\approx 91.66672$

6. 6

   Take the integer part. Since $91<a_{19}<92$, the integer part is $91$. A direct iteration of the recurrence in exact fractions reproduces $a_{19}=\tfrac{3003735}{32768}$, confirming the closed form.