Question

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Answer 1

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, The line segment is divided in 4 parts are :- a , (a + d) , (a + 2d) & (a + 3d). { where a is first term is d is common difference b/w the terms.}

So,

→ Fourth Term = Given = 14

→ T(4) = a + (n - 1)d = 14

→ a + (4 - 1)d = 14

→ a + 3d = 14 ------------------ Eqn.(1)

_____________________

Now, Also, given :-

→ 3rd part + 4th part = 3( first two parts.)

→ (a + 2d) + (a + 3d) = 3(a + a + d)

→ a + a + 2d + 3d = 3(2a + d)

→ 2a + 5d = 6a + 3d

→ 5d - 3d = 6a - 2a

→ 2d = 4a

→ d = 2a ------------------- Eqn.(2).

_____________________

Putting value of Eqn(2) in Eqn.(1) Now,

→ a + 3(2a) = 14

→ a + 6a = 14

→ 7a = 14

→ a = 2.

Hence,

→ Total Length of Line segment :-

=> a + (a + d) + (a + 2d) + (a + 3d)

=> a + (a + 2a) + (a + 4a) + (a + 6a)

=> a + 3a + 5a + 7a

=> 16a

=> 16 * 2

=> 32 (Ans.)

_________________________

Answer 2

${\huge{\bf{\red{\underline{Question:}}}}}$

A line segment divided into four parts forming an Arithmetic progression .The sum of length of third and fourth part is three times the sum of length of first two part.If the length fourth part is 14 cm. find the total length of line segment?

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Formula\: Used:}}}}$

$\dagger \: \boxed {\sf{ a_{n} = a + (n - 1)d}}$

${\bf{\blue{\underline{To\: Find:}}}}$

• Total length of line segment

${\bf{\blue{\underline{Now:}}}}$

Ist term,

$: \implies{\sf{ a_{1} = a + (1 - 1)d}}$

$: \implies{\sf{ a_{1} = a }}$

2nd term,

$\implies \: {\sf{ a_{2} = a + (n - 2)d}}$

$\implies \: {\sf{ a_{2} = a + d}}$

3rd term,

$\implies \: {\sf{ a_{3} = a + (3 - 1)d}}$

$\implies \: {\sf{ a_{3} = a + 2d}}$

4th term,

$\implies \: {\sf{ a_{4} = a + (4 - 1)d}}$

$\implies \: {\sf{ a_{4} = a + 3d}}$

It is given that 4th term is equal to 14,

$\implies \: {\sf{ a + 3d =14.....(1)}}$

It is also given that the 3rd and fourth term is the sum of the length of the first 2 part.

Therefore,

$: \implies{\sf{ (a + 2d)+(a + 3d) = 3 \times (a )+ (a + d)}}$

$: \implies{\sf{ a + 2d+a + 3d= 3 \times( a + a + d)}}$

$: \implies{\sf{ 2a + 5d= 3 \times( 2a + d)}}$

$: \implies{\sf{ 2a+5d= 6a + 3d}}$

$: \implies{\sf{ 5d-3d= 6a -2a}}$

$: \implies{\sf{ 2d= 4a}}$

$: \implies{\sf{ d= \frac{4}{2} a}}$

$: \implies \boxed{\sf{ d= 2 a......(2)}}$

Put value of d in (1),

$: \implies{\sf{ a + 3(2a)= 14}}$

$: \implies{\sf{ a + 6a= 14}}$

$: \implies{\sf{ 7a= 14}}$

$: \implies{\sf{ a= \frac{14}{7} }}$

$: \implies \boxed{\sf{ a= 2 }}$

Now put the value of d in (2),

$: \implies{\sf{ d= 2(2) }}$

$: \implies \boxed{\sf{ d= 4 }}$

___________________________________

• Total length of the line segment= a+a+da+2d+a+3d

$: \implies {\sf{ 2 + 2 + 4 + 2 + 2(4) + 2 + 3(4)}}$

$: \implies {\sf{ 12 + 2(4) + 3(4)}}$

$: \implies {\sf{ 12 + 8 + 12}}$

$: \implies {\sf{ 32}}$

$\dagger \: \: \: \boxed{\sf{ { \purple{Total \: length \: of \: the \: line \: segment \: = 32cm}}}}$