# Arithmetical progression

A lady who is recovering from a bike accident is beginning a training program to strengthen her injured leg. In the first day she walks only 50m. Everyday she walks 100m more than in the day before. Find how many meters does she walk during the first 30 days.

So...

1st day = 50 m 29 remaining days = 100 x 29 = 2900

2900 + 50 = 2950 m

Is this that easy?

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See this prior question on summing arithmetic progressions for conceptual insight. – Bill Dubuque Dec 31 '11 at 4:13

No, the first day she walks 50 m, then 150 m, then 250 m, and so on. You have 50, 100, 100, 100, etc.

So it is $50*30 + (0+100+200+\ldots +2900)$. This is an arithmetic progression as you say.

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As you noted, on the $30$-th day the lady walked a distance of $2950$ metres.

You were not told the full story. The lady lives at A, and her lover lives at B, which is $3000$ metres from A. On Day $1$, the lady walked $50$ metres, and her lover carried her the remaining $2950$ metres to B. On Day $2$, the lady walked $150$ metres, and was carried the remaining $2850$ metres to B. On Day $3$, the lady walked $250$ metres, and was carried $2750$ metres. And so on. On Day $30$, the lady walked $2950$ metres and was carried $50$ metres.

So the distances she walked went $50, 150, 250, \dots, 2750, 2850, 2950$, for a total of $$50+150+250+ \cdots +2750+ +2850+2950.$$ We could simply use a calculator to add up. But there is a better way. The distances she was carried went $2950, 2850, 2750, \dots, 250, 150, 50$. The total distance she was carried (we are adding up backwards) is therefore $$50+150+250+ \cdots +2750+ 2850+2950,$$ which is the same as the total distance she walked.

Every day, the distance walked plus the distance carried is $3000$, so in $30$ days, total distance walked plus total distance carried is $(30)(3000)$. But total distance walked is equal to the total distance carried, so the total distance walked is $$\frac{1}{2}(30)(3000).$$

Comment: Suppose that she walks a distance of $a$ the first day, and adds distance $d$ to her workout every day. Suppose also that she does this a total of $n$ days. Then on the last day she walks $a+(n-1)d$. Same story, except lady and lover live $2a+(n-1)d$ apart. The same argument shows that $$a+(a+d)+(a+2d)+\cdots +(a+(n-1)d)=\frac{1}{2}(n)(2a+(n-1)d).$$ Alternately, if the distance walked on Day $1$ is $a_1$, and increases by the same amount each day, reaching distance $a_n$ on Day $n$, then the total distance walked is $$\frac{1}{2}(n)(a_1+a_n).$$

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See here for much more on Gauss's method of summing an arithmetic progression via reflections. – Bill Dubuque Dec 31 '11 at 4:11

2950 m is how much she walked in the last day. The question asks for the total distance she walked in those 30 days.

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