SOLUTION: The salary scale for a chemical officer start at #1.1 million per annum. A rise of #72000 is given at the end of each year. Find the total amount of money earned in 12 years.

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Question 1204213: The salary scale for a chemical officer start at #1.1 million per annum. A rise of #72000 is given at the end of each year. Find the total amount of money earned in 12 years.
Found 3 solutions by MathLover1, mananth, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
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so, we need to find the sum of the first 12 terms of an arithmetic progression
with the first term a%5B1%5D=1100000 and the common difference d=+72000
Use the formula for the sum of n first terms of an AP


S%5Bn%5D+=+%28a%5B1%5D+%2B+%28%28n-1%29%2Ad%29%2F2%29%2An
in n=12 years will be
S%5B12%5D+=+%281100000%2B+%28%2812-1%29%2A72000%29%2F2%29%2A12
S%5B12%5D+=+17952000


Answer by mananth(16946) About Me  (Show Source):
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The salary scale for a chemical officer start at #1.1 million per annum. A rise of #72000 is given at the end of each year. Find the total amount of money earned in 12 years.
The sum of n terms of an Arithmetic sequence
Sn = n/2(2a+(n-1)d)
a=1100 000
d=72000
n=12
Sn= n/2(2a+(n-1)d
= 12/2 ( 2200000+11*72000)
=6*(2200000+792000)
17952000
Total amount of money earned by the chemical officer in 12 years is 17,952,000

Answer by greenestamps(13209) About Me  (Show Source):
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I am not fond of the formula generally shown in references for the sum of n terms of an arithmetic sequence. It is a good formal algebraic formula... but it is clumsy, and not very good at helping you learn about sums of arithmetic sequences.

In my opinion, an informal formula for finding the sum of an arithmetic sequence has much more educational value. The concept is simple:

sum = (number of terms) times (average of terms)

Finding the number of terms is nearly always easy; and in an arithmetic sequence, because of the equal spacing between terms, the average of all the terms is the average of the first and last terms. So we can write the formula as

sum = (number of terms) times (average of first and last terms)

In this problem, it is clear that there are 12 terms.

The first term is 1,100,000; the last term is 1,100,000 plus the common difference of 72,000 11 times: 1,100,000+792,000 = 1,892,000. The average of the terms is then (1,100,000+1,892,000)/2 = 1,496,000.

So the total earned in 12 years is

(12)(1,496,000)=17,592,000

ANSWER: #17,952,000