Upon graduation from Harvard, a graduate joined a consulting firm where she gained
several years of experience and an attractive compensation package. It is known that her salary increased every year, following an arithmetic progression, and that it took 13 years for
her salary to double that in her first year. Given that the graduate's total earnings for her first 10 years with the firm was $1,650,000, find her starting salary and annual increase.
please help me with this math. show eqns too please?
2016-05-26 4:09 am
回答 (1)
2016-05-26 4:43 am
The doubling time tells you the annual increase is 1/13 of her first year's wage. Let that first year's wage be represented by w.
The sum of 10 years' wages is then
.. 1.65M = 10w + (10*9)/2*w/13 ... from the formula for the sum of an arithmetic sequence
.. 1.65M = w(10+45/13) = 175w/13
.. 1650000*13/175 = w ≈ 122,571.43
And the annual increase is
.. 122,571.43/13 ≈ 9428.57
Her starting salary was $122,571.43, and her annual increase was $9,428.57.
_____
The sum of the first n terms of a sequence whose n-th term is
.. a_n = a + (n-1)d
is
.. sum = n*a + n*(n-1)*d/2
The sum of 10 years' wages is then
.. 1.65M = 10w + (10*9)/2*w/13 ... from the formula for the sum of an arithmetic sequence
.. 1.65M = w(10+45/13) = 175w/13
.. 1650000*13/175 = w ≈ 122,571.43
And the annual increase is
.. 122,571.43/13 ≈ 9428.57
Her starting salary was $122,571.43, and her annual increase was $9,428.57.
_____
The sum of the first n terms of a sequence whose n-th term is
.. a_n = a + (n-1)d
is
.. sum = n*a + n*(n-1)*d/2
收錄日期: 2021-04-21 18:45:36
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