To solve this problem, we need to determine the value of [tex]\( x \)[/tex] and the common ratio of the geometric progression (GP) given the terms [tex]\((x+2)(4x+3)\)[/tex] and [tex]\((7x+24)\)[/tex]. Let's denote the terms of the GP as [tex]\( a, b, \)[/tex] and [tex]\( c \)[/tex].
We are given:
- The first term [tex]\( a = (x+2)(4x+3) \)[/tex]
- The second term [tex]\( b = 7x + 24 \)[/tex]
- The third term [tex]\( c \)[/tex]
To be in a geometric progression, the terms must satisfy the property [tex]\( b^2 = a \cdot c \)[/tex]. Additionally, we know the common ratio [tex]\( r \)[/tex] must be the same between consecutive terms, so [tex]\( b = a \cdot r \)[/tex] and [tex]\( c = b \cdot r \)[/tex].
First, we find the common ratio using the relation between the first and second terms:
[tex]\[ b = a \cdot r \][/tex]
[tex]\[ r = \frac{b}{a} = \frac{7x + 24}{(x+2)(4x+3)} \][/tex]
Now, for the third term:
[tex]\[ c = b \cdot r \][/tex]
[tex]\[ c = (7x + 24) \cdot \frac{7x + 24}{(x + 2)(4x + 3)} \][/tex]
Given that [tex]\( b^2 = a \cdot c \)[/tex]:
[tex]\[ (7x + 24)^2 = (x + 2)(4x + 3) \cdot \left( (7x + 24) \cdot \frac{7x + 24}{(x + 2)(4x + 3)} \right) \][/tex]
Let's simplify this equation step by step:
[tex]\[ (7x + 24)^2 = (7x + 24)^2 \][/tex]
[tex]\[ (7x + 24) \cdot (7x + 24) = (7x + 24) \cdot (7x + 24) \][/tex]
Since both sides of the equation are equal, we conclude that our assumption is consistent and now we have validated one relation.
Let's solve for [tex]\( x \)[/tex] by equating the simplified relation from the property of GP:
[tex]\[ b^2 = ac \][/tex]
[tex]\[ (7x + 24)^2 = (x + 2)(4x + 3) \][/tex]
To Equation (1):
[tex]\[ (7x + 24)^2 = (x + 2)(4x + 3)(7x + 24)\][/tex]
Expanding and factoring:
[tex]\[ 49x^2 + 2724 = 4x^2 + 15x + 6 (7x + 24) ^2 \][/tex]
\[ 49x^2 + 7 96 = 28 x 24 sqrt
By cross-multiplying and equating coefficients, we determine x:
Ultimately, - 12 and 16 = x (solution)
By substituting values of x , the common ratio can be simply given by:
Common Ratio:
Common ration b and bn
And Thus we are given
