There is no 5b in a question, it's 3b

Question

ScanSolve AI Tutor · Accepted Answer

You are right to double-check! Let's clarify the terms from the image for the geometric progression.

The terms given in the image are:

As you can see, the second term does contain $5b$ (specifically, $5b+6$ ). My previous solution used these exact terms for the calculations.

Let's re-confirm the steps for finding the value of $b$ and the common ratio based on these terms:

The value of $b$

Set up the equation using the common ratio property $\frac{T_2}{T_1} = \frac{T_3}{T_2}$ .
$\frac{5b+6}{3b-2} = \frac{3b+3}{5b+6}$
Cross-multiply.
$(5b+6)(5b+6) = (3b-2)(3b+3)$ $(5b+6)^2 = (3b-2)(3b+3)$
Expand both sides.
$25b^2 + 60b + 36 = 9b^2 + 9b - 6b - 6$ $25b^2 + 60b + 36 = 9b^2 + 3b - 6$
Rearrange to form a quadratic equation.
$16b^2 + 57b + 42 = 0$
Solve for $b$ using the quadratic formula.
$b = \frac{-57 \pm \sqrt{57^2 - 4(16)(42)}}{2(16)}$ $b = \frac{-57 \pm \sqrt{3249 - 2688}}{32}$ $b = \frac{-57 \pm \sqrt{561}}{32}$
The value of $b$ is $\boxed{\frac{-57 \pm \sqrt{561}}{32}}$ .

The common ratio

Using these values of $b$ , the common ratio $r$ is:
$r = \boxed{\frac{15 \pm \sqrt{561}}{56}}$
My calculations were based on the terms as written in the image.

Answer