The first three terms of goemetric progression are 3b−2,3b+6 and 3b+3.find the value of b , and common ratio

Question

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Here are the solutions for Question 3 with the corrected terms.

The first three terms of the geometric progression are $T_1 = 3b-2$ , $T_2 = 3b+6$ , and $T_3 = 3b+3$ .
For a geometric progression, the ratio of consecutive terms is constant. This means $\frac{T_2}{T_1} = \frac{T_3}{T_2}$ .

The value of $b$

Set up the equation using the common ratio property.
$\frac{3b+6}{3b-2} = \frac{3b+3}{3b+6}$
Cross-multiply to eliminate the denominators.
$(3b+6)(3b+6) = (3b-2)(3b+3)$ $(3b+6)^2 = (3b-2)(3b+3)$
Expand both sides of the equation.
$9b^2 + 36b + 36 = 9b^2 + 9b - 6b - 6$ $9b^2 + 36b + 36 = 9b^2 + 3b - 6$
Rearrange the terms to form a linear equation.
Notice that the $9b^2$ terms cancel out.
$9b^2 - 9b^2 + 36b - 3b + 36 + 6 = 0$ $33b + 42 = 0$
Solve for $b$ .
$33b = -42$ $b = \frac{-42}{33}$ $b = -\frac{14}{11}$
The value of $b$ is $\boxed{-\frac{14}{11}}$ .

The common ratio
Substitute the value of $b = -\frac{14}{11}$ into the expressions for the terms.
First term $T_1 = 3b-2 = 3\left(-\frac{14}{11}\right) - 2 = -\frac{42}{11} - \frac{22}{11} = -\frac{64}{11}$ .
Second term $T_2 = 3b+6 = 3\left(-\frac{14}{11}\right) + 6 = -\frac{42}{11} + \frac{66}{11} = \frac{24}{11}$ .
Calculate the common ratio $r = \frac{T_2}{T_1}$ .
$r = \frac{\frac{24}{11}}{-\frac{64}{11}}$ $r = \frac{24}{11} \times \left(-\frac{11}{64}\right)$ $r = -\frac{24}{64}$ $r = -\frac{3}{8}$
The common ratio is $\boxed{-\frac{3}{8}}$ .

The first three terms of goemetric progression are 3b-2,3b+6 and 3b+3.find the value of b , and common ratio

Answer

Step-by-step solution

Set up the equation using the common ratio property.

Cross-multiply to eliminate the denominators.

Expand both sides of the equation.

Rearrange the terms to form a linear equation.

Solve for $b$ .

Substitute the value of $b = -\frac{14}{11}$ into the expressions for the terms.

Calculate the common ratio $r = \frac{T_2}{T_1}$ .

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