Antonio and Barbara start work at the same company on the same day. They each earn an annual salary of \(8000\) euros during the first year of employment. The company gives them a salary increase following the completion of each year of employment. Antonio is paid using plan A and Barbara is paid using plan B.

Plan A: The annual salary increases by \(450\) euros each year.

Plan B: The annual salary increases by \(5\,\% \) each year.

Calculate

i)     Antonio’s annual salary during his second year of employment;

ii)    Barbara’s annual salary during her second year of employment.

Write down an expression for

i)     Antonio’s annual salary during his \(n\) th year of employment;

ii)    Barbara’s annual salary during her \(n\) th year of employment.

Determine the number of years for which Antonio’s annual salary is greater than or equal to Barbara’s annual salary.

Both Antonio and Barbara plan to work at the company for a total of \(15\) years.

i)     Calculate the total amount that Barbara will be paid during these \(15\) years.

ii)    Determine whether Antonio earns more than Barbara during these \(15\) years.

i)     \(8450\,({\text{euro}})\)       (A1)

ii)    \(8000 \times 1.05\)       (M1)

Note: Award (M1) for \(8000 \times 1.05\)  OR \(\left( {8000 \times 0.05} \right) + 8000.\)

\( = 8400\,({\text{euro}})\)       (A1)(G3)

i)     \(8000 + 450\,\left( {n - 1} \right)\,\,\,\left( {{\text{accept}}\,\,450\,n + 7550} \right)\)        (M1)(A1)

Note: Award (M1) for substitution in arithmetic sequence formula; (A1) for correct substitutions.

ii)    \(8000 \times {1.05^{\left( {n - 1} \right)\,}}\)        (M1)(A1)

Note: Award (M1) for substitution in arithmetic sequence formula; (A1) for correct substitutions.

\(8000 + 450\,(n - 1) \geqslant 8000 \times {1.05^{n - 1}}\)        (M1)

Note: Award (M1) for setting a correct inequality using their expressions for (b)(i) and (b)(ii). Accept an equation.

OR

list of at least 4 correct terms of each sequence        (M1)

Note: Award (M1) for correct lists corresponding to their answers for parts (b)(i) and (b)(ii).

\(6\)        (A1)(ft)(G2)

Note: Value must be an integer for the final (A1) to be awarded. Follow through from parts (b)(i) and (b)(ii). Award (G1) for a final answer of \(6.70018...\) seen without working.

i)     \({S_{15}} = \frac{{8000 \times \left( {{{1.05}^{15}} - 1} \right)}}{{1.05 - 1}}\)         (M1)(A1)(ft)

Note: Award (M1) for substitution into geometric series formula and (A1) for correct substitution of \({u_1}\) and their \(r\) from part (b)(ii). Follow through from part (b)(ii).

OR

\(8000 + 8400 + 8820... + 15839.45\)        (M1)(A1)(ft)

Note: Follow through from part (b)(ii).

\( = 173\,000\,({\text{euro}})\,\,\,(172629...)\)         (A1)(ft)(G2)

ii)    \({S_{15}} = \frac{{15}}{2}\left( {2 \times 8000 + 450 \times 14} \right)\)         (M1)(A1)(ft)

Note: Award (M1) for substitution into arithmetic series formula and (A1) for correct substitution, using their first term and their last term from part (b)(i), or their \({u_1}\) and \(d\). Follow through from part (b)(i).

OR

\(8000 + 8450 + 8900... + 14300\)          (M1)(A1)(ft)

Note: Follow through from part (b)(i).

\( = 167\,000\,({\text{euro}})\,\,\,(167\,250)\)         (A1)(ft)(G2)

Antonio does not earn more than Barbara

(his total salary will be less than Barbara’s)         (A1)(ft)

Note: Award (A1)(ft) for a final answer that is consistent with their part (d)(i) and (d)(ii). Accept “Barbara earns more”. The final (A1) can only be awarded if two total salaries are seen.

Question 5: Arithmetic and Geometric progression
Most candidates calculated the salaries in the second year correctly. The most common error was to calculate the salaries for the third instead of the second year. In part (b) the use of \(n\) instead of \(n - 1\)  was very common. For the geometric sequence often a ratio of 0.05 instead of 1.05 was used. Also many of the expressions given did not represent a geometric sequence. Candidates who used a list for part (c) did usually better than the ones that tried to solve an equation. In part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression and not the series. In general this question part was not answered well.

Question 5: Arithmetic and Geometric progression

Most candidates calculated the salaries in the second year correctly. The most common error was to calculate the salaries for the third instead of the second year. In part (b) the use of \(n\) instead of \(n - 1\)  was very common. For the geometric sequence often a ratio of 0.05 instead of 1.05 was used. Also many of the expressions given did not represent a geometric sequence. Candidates who used a list for part (c) did usually better than the ones that tried to solve an equation. In part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression and not the series. In general this question part was not answered well.

Question 5: Arithmetic and Geometric progression

Most candidates calculated the salaries in the second year correctly. The most common error was to calculate the salaries for the third instead of the second year. In part (b) the use of \(n\) instead of \(n - 1\)  was very common. For the geometric sequence often a ratio of 0.05 instead of 1.05 was used. Also many of the expressions given did not represent a geometric sequence. Candidates who used a list for part (c) did usually better than the ones that tried to solve an equation. In part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression and not the series. In general this question part was not answered well.

Question 5: Arithmetic and Geometric progression

Most candidates calculated the salaries in the second year correctly. The most common error was to calculate the salaries for the third instead of the second year. In part (b) the use of \(n\) instead of \(n - 1\)  was very common. For the geometric sequence often a ratio of 0.05 instead of 1.05 was used. Also many of the expressions given did not represent a geometric sequence. Candidates who used a list for part (c) did usually better than the ones that tried to solve an equation. In part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression and not the series. In general this question part was not answered well.