# Arithmetic sequence involving years

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1. May 24, 2016

1. The problem statement, all variables and given/known data
A 25 year old programme for building new houses began in Core Town in the year 1986 and finished in 2010.

The number of houses built each year form an arithmetic sequence. Given that 238 houses were built in 2000 and 108 in 2010, find the number of houses built in 1986

2. Relevant equations
$U_n = a + (n - 1)d$

3. The attempt at a solution
$d = \frac{108 - 238}{10} = -13$
$238 = a + ((2000- 1986) - 1) * -13$
$a = 407$

But the real answer is $a=420$ by adding 1 to the 2000-1968. I do not understand why.
Can someone enlighten me?

Last edited: May 24, 2016
2. May 24, 2016

### rpthomps

There's a mistake in your work above.

3. May 24, 2016

Thanks. Fixed, but the question remains

4. May 24, 2016

### SammyS

Staff Emeritus
Do you mean that
The number of houses built each year form an arithmetic progression (sequence).​
?

Define what is meant by n .

5. May 24, 2016

Fixed. Thanks.
n is the number of years starting from 1986

6. May 24, 2016

### SammyS

Staff Emeritus
So, is 1986 year n=1 or is it year n=0 ?

7. May 24, 2016

n=1

8. May 24, 2016

### SammyS

Staff Emeritus
In that case, n = year - 1985

9. May 24, 2016

Can you explain? I do not understand
I have exams in half an hour D:

10. May 24, 2016

### SammyS

Staff Emeritus
In other words, for the year 2000, n = 2000 -1985 = 15

11. May 25, 2016

Thank god it did not come in exam

I know but why 1985 instead of 1986? The question says 1986 is the starting year

12. May 25, 2016

### SammyS

Staff Emeritus
Well, as you said, for the year 1986 n should be 1. 1986 - 1986 = 0, not 1.

1987 should give n = 2, right? But 1987 - 1986 = 1 not 2. Etc.

13. May 25, 2016