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October 1, 2014

October 1, 2014

Posted by **Misty** on Tuesday, June 26, 2012 at 5:58pm.

- math -
**Reiny**, Tuesday, June 26, 2012 at 7:45pmlet's look at the pattern

stage 1

they are 42 miles apart

let the time they meet be after t hrs.

distance travelled by boy 2 = 7t

distance travelled by pigeon = 21t

21t + 7t = 42

28t = 42,

t = 1.5

so the pigeon has travelled 21t or 21(1.5) miles

Stage 2

they are 21 miles apart, ( each boy has gone 7(1.5) miles)

define t the same way as before

distance travelled by boy 1 = 7t

distance travelled by pigeon = 21t

21t+7t=21

t = .75

so the pigeon has travelled 21(1.5) + 21(.75) miles

stage 3

they are 10.5 miles apart (each boy has gone 7(.75) miles, leaving 21 - 2(7)(.75) = 10.5)

distance travelled by boy 2 was 7t

distance travelled by pidgeon was 21t

21t+7t=10.5

t = .375

so the pigeon so far has flown 21(1.5) + 21(.75) + 21(.375)

etc ....

so the total distance covered by the pigeon is

21(1.5) + 21(.75) + 21(.375) + ... to infinity

= 21(1.5 + .75 + .375) + ..

the part inside the bracket is an infinite series with

a = 21.5 , and r = .5

sum = 21 ( 1.5/(1-.5) ) = 21(1.5/.5) = 21(3) = 63 miles

- math -
**Steve**, Tuesday, June 26, 2012 at 11:00pmOr, since the boys live 42 miles apart, and they each ride at 7mph, they ride 21 miles each, in 3 hours.

The pigeon, flying at 21 mph for those 3 hours, covers 63 miles.

- math -
**Reiny**, Tuesday, June 26, 2012 at 11:43pmSteve, Of course!!!!

How very clever of you

- math -
**Steve**, Wednesday, June 27, 2012 at 10:12amaw, shucks. > scuff scuff <

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