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.

Answer 1

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

Answer 2

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

...