Question

a line segment is divided into four parts forming an arithmetic progression the sum of the lengths of third and fourth parts is three times the sum of the lengths of first two parts if the length of 4th part is 14 cm find the total length of the line segment​

Answer 1

Answer:

The total length of the line segment = 32 cm

Step-by-step explanation:

Consider the first term is $a$ and a common difference is $d$,

Then the $\text{n}^{\text{th}}\text{ term}=a+(n-1)d$

the ${1}^{\text{st}}\text{ term}=a$

the ${2}^{\text{nd}}\text{ term}=a+(2-1)d=a+d$

the ${3}^{\text{rd}}\text{ term}=a+(3-1)d=a+2d$

the ${4}^{\text{th}}\text{ term}=a+(4-1)d=a+3d$

As per problem,

${4}^{\text{th}}\text{ term}=a+(4-1)d=a+3d=14....................\text{Equation-1}$

And

$(a+2d)+(a+3d)=3\times[a+(a+d)]\\\\\Rightarrow(a+2d+a+3d)=3\times(2a+d)\\\\\Rightarrow(2a+5d)=(6a+3d)\\\\\Rightarrow(2a+5d-6a-3d)=0\\\\\Rightarrow(2d-4a)=0\\\\\Rightarrow(4a-2d)=0\text{ .....................Equation-2}$

Multiply the equation-1 with 4 and then subtract equation-2 from equation 1 we get,

$(4a+12d)-(4a-2d)=54-0\\\\(4a+12d-4a+2d)=54\\\\14d=54\\\\d=4$

Put the value of d in equation -1 we get,

$a+3d=14\\\\a+3\times4=14\\\\a=14-12\\\\a=2$

Therefore the first term of the series $a=2$ and the difference is $d=4$

No of term $n=4$

therefore the Sum of Arithmetic series is

$S_{n}=\frac{n}{2}[2a+(n-1)d]\\\\\Rightarrow{S}_{n}=\frac{4}{2}[2\times2+(4-1)4]\\\\\Rightarrow{S}_{n}=2\times[4+12]\\\\\Rightarrow{S}_{n}=2\times16\\\\\Rightarrow{S}_{n}=32$

Therefore the total length of the line segment = 32 cm

Answer 2

The total length of the line segment​ is 32 cm.

Step-by-step explanation:

We are given that a line segment is divided into four parts forming an arithmetic progression the sum of the lengths of third and fourth parts is three times the sum of the lengths of first two parts.

Also, the length of 4th part is 14 cm.

Let the four parts forming an arithmetic progression be $a_1, a_2, a_3 \text { and } a_4$.

Now, the sum of n terms of an A.P. is given by;

           $a_n=a + (n-1) d$

where, a = first term of an AP  

            d = common difference

So,  $a_1 =a$

$a_2 = a + (2 -1) d = a + d$

$a_3 = a + (3 -1) d = a + 2d$

$a_4 = a + (4 -1) d = a + 3d$

Now, according to the question;

• The First condition states that the sum of the lengths of third and fourth parts is three times the sum of the lengths of first two parts, that means;

                  $(a+2d) + (a+ 3d) = 3 \times (a + (a+d))$

                              $2a + 5d = 3(2a +d)$

                               $2a + 5d=6a + 3d$

                                $6a - 2a = 5d - 3d$

                                  $4a = 2d$

                                    d = 2a ------------- [Equation 1]

• The second condition states that the length of 4th part is 14 cm, that means;

                             a + 3d = 14

                             a + 3(2a) = 14     {using equation 1}

                             a + 6a = 14

                                a = $\frac{14}{7}$ = 2

Now, putting value of a in equation 1 we get;

                           d = 2a = 2 $\times$ 2 = 4

SO, the total length of the line segment​ = (a) + (a + d) + (a + 2d) + (a + 3d)

                           =  4 + (2 + 4) + (2 + 8) + (2 + 12)

                           =  4 + 6 + 10 + 14 = 32 cm.