Hello!
We have the following data:
a1 (first term or first year salary) = 32000
r (ratio or annual increase) = 300
n (number of terms or each year worked)
We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:
formula:
[tex] a_n = a_1 + (n-1)*r [/tex]
* second year salary
[tex] a_2 = a_1 + (2-1)*300 [/tex]
[tex] a_2 = 32000 + 1*300 [/tex]
[tex] a_2 = 32000 + 300 [/tex]
[tex] \boxed{a_2 = 32300} [/tex]
* third year salary
[tex] a_3 = a_1 + (3-1)*300 [/tex]
[tex] a_3 = 32000 + 2*300 [/tex]
[tex] a_3 = 32000 + 600 [/tex]
[tex] \boxed{a_3 = 32600} [/tex]
* fourth year salary
[tex] a_4 = a_1 + (4-1)*300 [/tex]
[tex] a_4 = 32000 + 3*300 [/tex]
[tex] a_4 = 32000 + 900 [/tex]
[tex] \boxed{a_4 = 32900} [/tex]
* fifth year salary
[tex] a_5 = a_1 + (5-1)*300 [/tex]
[tex] a_5 = 32000 + 4*300 [/tex]
[tex] a_5 = 32000 + 1200 [/tex]
[tex] \boxed{a_5 = 33200} [/tex]
We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:
* sixth year salary
[tex] a_6 = a_1 + (6-1)*300 [/tex]
[tex] a_6 = 32000 + 5*300 [/tex]
[tex] a_6 = 32000 + 1500 [/tex]
[tex] \boxed{a_6 = 33500} [/tex]
* seventh year salary
[tex] a_7 = a_1 + (7-1)*300 [/tex]
[tex] a_7 = 32000 + 6*300 [/tex]
[tex] a_7 = 32000 + 1800 [/tex]
[tex] \boxed{a_7 = 33800} [/tex]
* eighth year salary
[tex] a_8 = a_1 + (8-1)*300 [/tex]
[tex] a_8 = 32000 + 7*300 [/tex]
[tex] a_8 = 32000 + 2100 [/tex]
[tex] \boxed{a_8 = 34100} [/tex]
* ninth year salary
[tex] a_9 = a_1 + (9-1)*300 [/tex]
[tex] a_9 = 32000 + 8*300 [/tex]
[tex] a_9 = 32000 + 2400 [/tex]
[tex] \boxed{a_9 = 34400} [/tex]
* tenth year salary
[tex] a_{10} = a_1 + (10-1)*300 [/tex]
[tex] a_{10} = 32000 + 9*300 [/tex]
[tex] a_{10} = 32000 + 2700 [/tex]
[tex] \boxed{a_{10} = 34700} [/tex]
we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:
* eleventh year salary
[tex] a_{11} = a_1 + (11-1)*300 [/tex]
[tex] a_{11} = 32000 + 10*300 [/tex]
[tex] a_{11} = 32000 + 3000 [/tex]
[tex] \boxed{\boxed{a_{11} = 35000}}\end{array}}\qquad\checkmark [/tex]
Respuesta:
In the eleventh year of salary he will earn more than 34700, in the case, this value will be 35000
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¡Espero haberte ayudado, saludos... DexteR! =)
