On a path there was an odd number of rocks with a distance of 10m between them, we want to put them all where the middle one is, if we start to collect from one of the ends and when we finish we have walked 3km in total, how many rocks are there?


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Question Number 182704 by HeferH last updated on 13/Dec/22

$O n a p a t h t h e r e w a s a n o d d n u m b e r o f r o c k s$ $w i t h a d i s t a n c e o f 10 m b e t w e e n t h e m, w e$ $w a n t t o p u t t h e m a l l w h e r e t h e m i d d l e o n e i s,$ $i f w e s t a r t t o c o l l e c t f r o m o n e o f t h e e n d s a n d$ $w h e n w e f i n i s h w e h a v e w a l k e d 3 k m i n t o t a l,$ $h o w m a n y r o c k s a r e t h e r e ?$
	Answered by TheSupreme last updated on 13/Dec/22

	$i f n = 1 S_{1} = 0$ $i f n = 3 S_{3} = 3 * 10$ $i f n = 5 S_{5} = S_{3} + 10 + 2 * 3 * 10$ $S_{2 n + 1} = S_{2 n - 1} + 10 * (n - 1) + 3 * n * 10$ $S_{2 n + 1} = \underset{i = 1}{\sum^{n}} 10 (i - 1) + 30 * i = \underset{i = 1}{\sum^{n}} 40 i - 10$ $S_{2 n + 1} = 40 \frac{n (n + 1)}{2} - 10 n$ $3000 = 20 n^{2} + 10 n = 10 n (2 n + 1)$ $n = 12$
	Answered by mr W last updated on 13/Dec/22

	$s a y t h e r e a r e 2 n + 1 r o c k s w i t h d i s t a n c e d$ $b e t w e e n t h e m .$ $t o t a l d i s t a n c e t o w a l k :$ $L = 4 (d + 2 d + 3 d + . . . + n d) - n d = n (2 n + 1) d$ $3000 = n (2 n + 1) 10$ $2 n^{2} + n - 300 = 0$ $\Rightarrow n = 12$ $i . e . t h e r e a r e 2 \times 12 + 1 = 25 r o c k s$
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