Question

find x if 3x^2 +4x +4, 2x^2 +3x +3 and 3x + 8 are the term of A.P where x is a natural number​

Answer 1

Answer:

Step-by-step explanation:
2x²+3x+3-(3x²+4x+4)=3x+8-(2x²+3x+3)
=>2x²+3x+3-3x²-4x-4=3x+8-2x²-3x-3
=>-x-x²-1=5-2x²
=>x²-x-6=0
=>x²+(3-2)x-6=0
=>x²+3x-2x-6=0
=>x(x+3)-2(x+3)=0
=>(x-2)(x+3)=0
Either, x=2
OR
x=-3

Answer 2

Given:

3x² + 4x + 4

2x² + 3x + 3

3x + 8  are the terms of the AP

To Find:

The value of x

Solution:

x+3, 3x, 3x+5 are the terms of the AP.

To find any series in AP,

Second term - First term = third term - second term

So,

$a_{2} - a_{1} = a_{3}- a_{2}$

substituting the terms,

(3x)-(x+3) = (3x + 5)-(3x)

Now simplifying the values,

⇒ 3x-3-x = 3x+5-3x

⇒ 2x - 3 = 5

⇒ 2x = 5 + 3

⇒ 2x = 8

⇒ x = 8/2

⇒ x= 4

Hence, the value of x is 4.

Therefore x is a natural number.