Question 1175621: A man buys premium bonds every year . In the first year , he buys sh.2000 worth of bonds . If every year he increases his annual investments in bonds by sh.600 , how long will he take for his total investment in bonds to be sh.37000
Found 2 solutions by greenestamps, Boreal:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
His investments for each year form an arithmetic sequence:
2000, 2600, 3200, 3800, 4400, 5000, 5600, 6200, 6800, ...
The fastest path to the answer is to add up the successive amounts until the total reaches 37000 or more. You will quickly find that his total investment over 8 years is less than 37000 and the total over 9 years is greater then 37000.
For an algebraic solution, using the formula for the sum of an arithmetic sequence....
first payment: 2000
payment in n-th year: 2000+600(n-1) = 1400+600n
Since the sequence is arithmetic, the sum of payments over n years is the number of years, multiplied by the average of the amounts from the first and last years:
We want that sum to be 37000:
That quadratic does not factor; you need to find the solution using the quadratic formula or a graphing calculator. Of those two options, clearly the graphing calculator is more efficient.
But you found the answer much faster by performing a few simple additions....
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! a1=2000
a2=3200
an=2000+(n-1)600
sum=(n/2)(2a+(n-1)d)
37000=(n/2)(4000+(n-1)600)
74000=n(4000+600n-600))=3400n+600n^2
600n^2+3400n-74000=0
6n^2+34n-740=0
3n^2+17n-370=0
n=(1/6)(-17+ sqrt (4729)); sqrt term=68.76
n=51.76/6 or 8.6 years
between the 8th and 9th years.
2000
2600
3200
3800
4400
sum for 5 years is 16000
then
5000 (21000)
5600 (26600)
6200 (32800)
6800 (39600) on year 9
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