Question

A piece of string of length 5m long is cut into n pieces in such a way that the lengths of the pieces are an arithmetic sequence. If the lengths of the longest and the shortest pieces are 1m and 25cm respectively, calculate n.

Answers

Bosnian
In an arithmetic progression:

an = a + ( n - 1 ) d

where

a = the initial term

d = the common difference of successive members

an = the nth term

Lengths shortest pieces:

a1 = a + ( 1 - 1 ) d = a + 0 ∙ d = a

a1 = a = 25 cm

Lengths longest pieces:

an = 1 m = 100 cm

an = a + ( n - 1 ) d

100 = 25 + ( n - 1 ) d

The sum of n terms of an arithmetic progression:

Sn = ( n / 2 ) [ 2 a + ( n -1 ) d ]

In this case a = 25 cm so:

Sn = ( n / 2 ) [ 2 ∙ 25 + ( n -1 ) d ]

Sn = ( n / 2 ) [ 50 + ( n -1 ) d ]

The sum of n terms of this arithmetic progression is 5 m

Sn = 5 m = 500 cm

500 = ( n / 2 ) [ 50 + ( n -1 ) d ]

Now you must solve system of two equations:

25 + ( n - 1 ) d = 100

( n / 2 ) [ 50 + ( n -1 ) d ] = 500

The solution is:

d = 75 / 7 , n = 8

Your arithmetic progression:

a1 = 25

a2 = 25 + 75 / 7 = 175 / 7 + 75 / 7 = 250 / 7

a3 = 250 / 7 + 75 / 7 = 325 / 7

a4 = 325 / 7 + 75 / 7 = 400 / 7

a5 = 400 / 7 + 75 / 7 = 475 / 7

a6 = 475 / 7 + 75 / 7 = 550 / 7

a7 = 550 / 7 + 75 / 7 = 625 / 7

a8 = 625 / 7 + 75 / 7 = 700 / 7 = 100

You can check the sum of this arithmetic progression.

a1 + a2 + a3 + a4 + a5 + a6 + a7 =

25 + 250 / 7 + 325 / 7 + 400 / 7 + 475 / 7 + 550 / 7 + 625 / 7 + 100 =

25 + ( 250 / 7 + 325 / 7 + 400 / 7 + 475 / 7 + 550 / 7 + 625 / 7 ) + 100 =

25 + 2625 / 7 + 100 = 25 + 375 + 100 = 500

The sum of this arithmetic progression = 500 cm
Bosnian
My little typo.

the sum of this arithmetic progression is:

a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 =

25 + 250 / 7 + 325 / 7 + 400 / 7 + 475 / 7 + 550 / 7 + 625 / 7 + 100

...

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