Stephanie M. answered • 05/26/15
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There are 52 weeks in a year. So, his first year of allowance, the boy got 6(52) = 312 cents. His second year of allowance, he got 7(52) = 364 cents. His third year, he got 8(52) = 416 cents. The pattern continues like this:
6(52), 7(52), 8(52), 9(52), 10(52), ...
312, 364, 416, 468, 520, ...
Notice that each time we're adding 52 cents. That means we've got an Arithmetic Sequence with a common difference of d = 52. (Hopefully you guys are learning about those right now in school.)
To figure out the total he would have been given, we want to find the sum of the sequence from the first term to the 45th term (when the man is 50 years old).
The equation for the partial sum of an Arithmetic Sequence is:
Sn = n(a1 + an)/2,
where n is the number of the last term you're interested in, a1 is the value of the first term, and an is the value of the last term. For your problem, n = 45. a1 = 312 (which we found earlier) and an = 50(52) = 2600. Plug those in:
S45 = 45(312 + 2600)/2 = 65520
This means that the man would have been given a grand total of 65520 cents, or $655.20.
Stephanie M.
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Hi Carol,
I see. You may want to re-post the question and include that detail. The clearer your question, the less likely you'll get an unhelpful answer!
Stephanie
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05/27/15
Carol N.
Thank you, Stephanie. Your answer was very helpful. I edited the question to try to find out, now, what amount of allowance would he be entitled to ....considering inflation.
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05/27/15
Carol N.
05/27/15