Key to Robinsons New University Algebra

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6. Let n= the number of days the first person travels, d=lj and l=:l ■^{n — l)d=n.
. «=q(1 + n) = the whole distance.
and 15(w-6)= « " hence, in{l+n) = 15(n-6\ or, n' + w=30n— 180, n'-29n=-180.
First person travels.
n = Y=t-V-= 9 or 20, -6 -6, Second « ** 3 or 14, Ans, Explanation. — Call the first person A, and the second B. Now B overtakes and passes A after A has traveled 9 days and B 3 days. But as A is increasing his rate one mile per day, he finally gains on B, and overtakes and passes hi
...m after A has traveled 20 days, and B 14. They are together after having trav- eled 45 and 201 miles.
6. Let x=: one of the equal payments.
At the given rate per cent $60 will amount to $61 at the end of 60 days. As the rate of interest is ^-^^^ of the principal for a day, the firet partial payment will amount to the second to ^+3loo-^» the third to ^+3lJo^, and the last to x.
Hence the sum of the partial payments is, Q0x-\-^j\^{59x-\-58x + 5lz. . . . +x), or by summing the series, 60^ + tVo^' Since the debt is to be canceled, we must have 60a- + T¥o^= 61, T259ar=7320, whence ^=^1t2J7» ^^' (292) ARITHMETICAL PROGRESSION.


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