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MaRussiya [10]
2 years ago
15

Sarah is a computer engineer and manager and works for a software company. She receives a

Mathematics
1 answer:
daser333 [38]2 years ago
8 0

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = a

Given that:

a_4=129\\a_{10}=207

Formula for n^{th} term of an Arithmetic Progression is given as:

a_n=a+(n-1)d

Where d will represent the number of projects increased every year.

and n is the year number.

a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

Subtracting (2) from (1):

78 = 6d\\\Rightarrow d =13

By equation (1):

129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

Each project pays $500

Earnings in the twelfth year = 233 \times 500 = $116500

Sum of an AP is given as:

S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 \times 1938 = $969000

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