Question

question:-

The value of x for which 3x,(x+11) and (4x+3) are the three consecutive terms of an AP, is​

Answer 1

$\underline{\underline{\huge{\blue{\tt{\textbf Answer :-}}}}}$

Given:-

• the three consecutive terms of an AP, is 3x,(x+11) and (4x+3)

To find:-

• the value of x=???

Solution:-

Here,

$a = 3x$

an= 4x + 3

$n = 3$

we have to find

common difference of an Ap:-

difference( d1) = a2-a1 =( x + 1) - (3x) = - 2x + 11

we know that-

$a_{n} = a + (n - 1) \times d \\ 4x + 3 = 3x + (3 - 1) \times d \\ 4x + 3 - 3x = 2d \\ x + 3 = 2d..(1)$

from equation (1)

⟼x+3=2d

⟼x+3=2(-2x+11)

⟼x-3= -4x+22

⟼x+4x= 19

⟼5x= 19

⟼x= 19/5

Answer 2

Answer:

your full solution

hope it's helpful.