Question

If (x+1),3x and (4x+2) are first three terms of an AP, then its 5th term is

 17

 19

 28

 27​

Answer 1

Answer:

as the terms are in AP series,so,

3x-(x+1)=(4x+2)-(3x)

=> 3x-x-1=4x+2-3x

=> 2x-1=x+2

=> 2x-x= 2+1

=> x= 3

so 1st term=x+1= 3+1=4

2nd term=3x=3×3=9

3rd term=4x+2=4×3+2=12+2=14

so 1st term=a=4

common difference=9-4=5

5th term=a+(5-1).d=4+4.5=4+20=24

Answer 2

Given : Here, (x+1), 3x and (4x+2) are first three terms of an AP.

To find : The 5th term of the sequence.

$\qquad$_________________

We know that, the difference between the first term and second term is equal to the difference between the second term and third term of the progression.

So, we get :

$\sf \qquad\dashrightarrow~ 3x-(x+1)=(4x+2)-3x$

$\sf \qquad\dashrightarrow~ 3x-x-1=4x-3x+2$

$\sf \qquad\dashrightarrow~ 2x-1=x+2$

$\sf \qquad\dashrightarrow~ 2x-x=2+1$

$\bf \qquad\dashrightarrow~ x = 3$

☀️ Therefore, x = 3 in (x+1), 3x and (4x+2).

$\qquad$_________________

Finding the common difference “d” :

Substituting x = 3 in the terms and finding the difference :

$\sf \qquad\dashrightarrow~ 3x-(x+1)=(4x+2)-3x$

$\sf \qquad\dashrightarrow~ 3(3)-(3+1)=4(3)+2-3(3)$

$\sf \qquad\dashrightarrow~ 9-4=12+2-9$

$\sf \qquad\dashrightarrow~ 5=5$

☀️ Therefore, the common difference between the terms is 5.

$\qquad$_________________

Finding the 5th term :

We have :

$\qquad$ ☆ a = x+1 = 4

$\qquad$ ☆ d = 5

$\qquad$ ☆ n = 5

Substituting the formula and finding :

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

$\sf \qquad\dashrightarrow~ a_n=4+(5-1)(5)$

$\sf \qquad\dashrightarrow~ a_n=4+(4)(5)$

$\sf \qquad\dashrightarrow~ a_n=4+20$

$\bf \qquad\dashrightarrow~a_n = 24$

☀️ Therefore, the 5th term is 24.