Geometric sequence

Peter works in a factory with a starting salary of $1 000 per month. If he has an annual increment of 10% of his salary, find
a) his salary per month at the nth year,
b) his total income for the first 10 years (to the nearest dollar)


a)
Starting monthly salary = T(1) = a = 1000
Monthly increment = 10%
Common ratio = r = (1 + 10%) = 1.1
Number of years = n

T(n)
= arⁿ⁻¹
= 1000 x (1.1)ⁿ⁻¹

His salary per month at the nth year = $1000 x (1.1)ⁿ⁻¹


b)
T(1) + T(2) + …… + T(n)
= a + ar + ar² + …… + arⁿ⁻¹
= a(rⁿ - 1)/(r - 1)
= 1000(1.1¹⁰ - 1)/(1.1 – 1)
= 1000(1.1¹⁰ - 1)/0.1

his total income for the first 10 years
= $[1000(1.1¹⁰ - 1)/0.1] x 12
= $191249 (to the nearest dollars)

Peter works in a factory with a starting salary
of $1 000 per month. If he has an annual increment of 10% of his salary, find
a) his salary per month at the nth year,
1000*(1+10%)^n

= 1000 * ( 1.1 ) ^n

b) his total income for the first 10 years (to the
nearest dollar)

1000 * ( 1.1 ) ^ 10

= 1000 * 2.5937424601

= 2593.7424601

~ 2594