Question

A piece of wire 216 m long is cut into 27 pieces whose lengths are in Arithmetic Progression.
The sum of the lengths of the 3 shortest pieces is 15 cm.
Calculate
(a) The length of the shortest piece,
(b) The sum of the lengths of the three longest pie

Answer 1

Step-by-step explanation:

As we know that an a.p can be written as

a,(a+d),(a+2d),(a+3d),.............(a+Nd)

Now

it is given that a 216 m log wire cuts into 27 pieces

I.e

a+(a+d)+(a+2d)+..............(a+26d) = 216

27a + 351d = 216 ----------> EQ 1

also we know that 3 shortest wires length is 15 cm

it means

a+(a+d)+(a+2d) = 15

3a + 3d = 15 ---------> EQ 2

from EQ 1 and 2

3a + 3d = 15

27a + 351d = 216

multiply EQ 1 by 9 to make "a" same

27a + 27d = 405

27a + 351d = 216

-324d = 189

d = - 189/324

d = -63/108

= -7/12

Now d = -7/12

to find value of a put the value of d in equation

27a + 27(-7/12)= 405