Arithmetic Sequence Word Problem with floor heights

Question

The $7$th floor of a building is $23$m above street level and $13$th floor is $41$m above street level. What is the height (above street level) of the first floor and what is the height of one floor?

My working out is this:

$t_7=23$

$t_{13}=41$

$t_n=t_1 + (n-1) d$

$23= t_1 + 6d (1)$

$41 = t_1 + 12d (2)$

$18= 6d$

$d=3$

Sub $3$ into equation $2$

$41=t_1 + 12(3)$

$41= t_1 + 36$

$5 = t_1$

Can someone verify if I am doing it correct?

Answer 1

Yes you are correct. The first floor's hight above street level is (in meters): $$a_1=5 $$

Also, with $d$ indicating d is the difference between terms of the arithmetic progression, $$d=3$$

You should be able to verify the answer using the equation for the height of any floor from street level in meters to be:

$$a_n=5+(n-1)(3) \tag1$$

You already know that $a_7=23$, so:

Using (1), with $n=7$

$$a_7= 5+ (7-1)(3)=5+18=23$$

Same goes for the $13th$ floor, its height above the street in meters is $41$ using (1) again:

$$a_{13}=5+(13-1)3=5+36=41$$

Note that this site has a consice summary of Arithmatic Progression Formulae.

Answer 2

Yes it is perfectly correct. The first floor is $5m$ above the street level above which every floor with a height of $3m$.