mathematical expression for arithmetic sequence with 3 variables where 1 of them is known

Question

This is from an A-level exam paper, so it's fairly elementary. I'd like the know how to express the following mathematically.

A construction programme began in 1986 and finished in 2010. The number of houses built each year form an arithmetic sequence, in year 2000, 238 houses are built. in year 2010, 108 houses are built.

My attempt to express this:

$$ a_{n-i} = a + (n-i)(d)$$ n=corresponding year, i=starting year

$$ a_n = f(n-i)$$

$$ f(2000-1986) = 238$$

$$f(2010-1986) = 108 $$

Can I leave $a$ and $d$ out like that?

Answer 1

Okay, notice the trend first. In $2000$ we had $238$ houses and in $2010$ we had $108$ houses, so in $10$ years, the number of houses construction decreased by $108-238=130$, which means roughly $13$ houses every year. Now $1986$ is $14$ years before $2000$ so, $14*13=42$, so in $1986$, we will have $238+182=420$ houses!

Answer 2

The general term of an arithmetic sequence with difference $d$ and first term $a_1$ is $a_n=a_1+(n-1)d$.

We have that $a_{15}$, corresponding to the year $2000$ is equal to $238$, and similarly that $a_{25}$, corresponding to the year $2010$ is equal to $108$, thus

$$a_{15}=a_1+(15-1)d=238\\a_{25}=a_1+(25-1)d=108$$

By solving this system you can directly compute $a_1, d$ and thus any $a_n$.

Answer 3

An arithmetic sequence is generally written like so: $a_i = di + a_0$ where d is the Common Difference and $a_0$ is the Initial Term.

Your problem gives you that in 2000 there were 238 houses built, but only 108 houses were built in 2010. So $d = (108 - 238) / (2010 - 2000) = -13$.

To compute $a_0$ we simply plug d into the equation along with one of the data points. So $238 = -13(2000 -1986) + a_0$. Solve to get $a_0 = 420$ (if I did it in my head right).

So the formula would be $a_i = -13i + 420$. For i, plug-in the difference in years between the year in question and the initial year (1986).