Peter Pedals rode his bike a total of 500 miles in five days. Each day he rode 10 more miles than he had ridden on the previous day. How many miles did Peter ride on just the fifth day?
+6248 \(miles_1 = M\\ miles_k = miles_1 + 10(k-1),~k=1,2,\dots 5\\ \sum \limits_{k=1}^5 miles_k = \sum \limits_{k=1}^5 M+10(k-1) = 500\\ 5M + 10\sum \limits_{k=0}^4 k = 500\\ 5M + 10\dfrac{(4)(4+1)}{2} = 500\\ 5M + 100 = 500\\ M=80\\ miles_5 = 80 + 10(4) = 120~miles \)
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+128460 Call the distance he bikes the first day = D
The distance he bikes the ffth day = D + 4(10) = D + 40
So
Sum = [ Distance on first day + Distance on fifth day] * (number of days / 2)
500 = [ D + D + 40 ] * (5/2)
500 = [ 2D + 40 ] * (5/2) multiply both sides by 2/5
200 = 2D + 40 subtract 40 from each side
160 = 2D divide both sides by 2
80 = D
So....on the fifth day he bikes 80 + 40 = 120 miles
