Answer:
a). Geometric sequence
b). ![S_{t} =\$37185(1+\frac{6}{100})^{(t-1)}](https://tex.z-dn.net/?f=S_%7Bt%7D%20%3D%5C%2437185%281%2B%5Cfrac%7B6%7D%7B100%7D%29%5E%7B%28t-1%29%7D)
c). $46945.21 will be the salary at the start of 5th year.
Step-by-step explanation:
My salary for the first year is $37185.
If I get an increase of 6% every year then the next year salary will be,
![S_{1}=\$37185(1+\frac{6}{100})](https://tex.z-dn.net/?f=S_%7B1%7D%3D%5C%2437185%281%2B%5Cfrac%7B6%7D%7B100%7D%29)
= $39416.1
Similarly in the second year my salary will be
![S_{2}=\$39416.1(1+\frac{6}{100})](https://tex.z-dn.net/?f=S_%7B2%7D%3D%5C%2439416.1%281%2B%5Cfrac%7B6%7D%7B100%7D%29)
= $41781.07
So the sequence becomes $37185, $39416.1 $41781.07.....
And there is a common ratio in each successive term 'r' = 1.06
a). Therefore, it's a geometric sequence.
b). Explicit formula for this sequence will be in the form of
![S_{t} =\$37185(1+\frac{6}{100})^{(t-1)}](https://tex.z-dn.net/?f=S_%7Bt%7D%20%3D%5C%2437185%281%2B%5Cfrac%7B6%7D%7B100%7D%29%5E%7B%28t-1%29%7D)
Where t = duration in years
c). Salary at the start of 5th year,
![S_{(5)}=\$37185(1+\frac{6}{100})^{5-1}](https://tex.z-dn.net/?f=S_%7B%285%29%7D%3D%5C%2437185%281%2B%5Cfrac%7B6%7D%7B100%7D%29%5E%7B5-1%7D)
![S_{(5)}=\$37185(1.06)^{4}](https://tex.z-dn.net/?f=S_%7B%285%29%7D%3D%5C%2437185%281.06%29%5E%7B4%7D)
= $46945.21