Two brothers, Eoin and Peter, began work in 2005 on starting salaries of €20,000 and €17,000 per annum, respectively - Leaving Cert Mathematics - Question Question 1 - 2013

Both brothers will earn the same amount in 2009.

An arithmetic sequence is a sequence in which the difference between any two successive terms is a constant. In this case, Eoin's and Peter's salaries increase by fixed amounts each year.

Yes, I agree with Eoin because a constant amount is added to his salary each year, which fits the definition of an arithmetic sequence.

Eoin's salary in the nth year can be represented by the formula:

$T_n = 20000 + (n - 1)500 = 19500 + 500n$

To find Eoin's salary in 2015:

2015 corresponds to n = 11.

Thus, substituting into the formula:

$T_{11} = 19500 + 500(11) = 25000€$

The total amount earned by Peter from the first year to the nth year is given by:

$S_n = \frac{n}{2} \left(2(17000) + (n-1)(1250)\right) = \frac{n}{2} \left(34000 + (n-1)1250\right)$

Using the formula, we find:

For n = 11 (2005 to 2015):

$S_{11} = \frac{11}{2} \left(34000 + 10 imes 1250 \right) = 255750€$

The graph shows Peter's salary increasing constantly throughout the year, but this is not true; his salary increases in steps at the end of each year.