Calculate the value of $$ m $$. The sum of the first three terms of a geometric sequence is $$ 31 \frac{1}{2} $$. The sum of the fourth, fifth and sixth term of the same sequence is $$ 3 \frac{15}{16} $$. Determine the value of $$ t $$ common ratio $$ (r) $$.

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To find the value of $$ m $$ and the common ratio $$ r $$ in a geometric sequence, we can use the formula for the sum of the first $$ n $$ terms of a geometric sequence:

$$ S_n = \frac{a(1 - r^n)}{1 - r} $$

where:
- $$ S_n $$ is the sum of the first $$ n $$ terms,
- $$ a $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the number of terms.

Given that the sum of the first three terms is $$ 31 \frac{1}{2} $$ and the sum of the fourth, fifth, and sixth terms is $$ 3 \frac{15}{16} $$, we can set up two equations using the formula for the sum of the first $$ n $$ terms:

1. For the first three terms:
$$ 31 \frac{1}{2} = \frac{a(1 - r^3)}{1 - r} $$

2. For the fourth, fifth, and sixth terms:
$$ 3 \frac{15}{16} = \frac{ar^3(1 - r^3)}{1 - r} $$

We can simplify these equations and solve for $$ a $$ and $$ r $$ to find the value of $$ m $$ and the common ratio $$ r $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}31\frac{1}{2}=\frac{a\left(1-r^{3}\right)}{1-r}\3\frac{15}{16}=\frac{ar^{3}\left(1-r^{3}\right)}{1-r}\end{array}\right.$$
- step1: Calculate:
  $$\left\{ \begin{array}{l}\frac{63}{2}=a\left(1+r+r^{2}\right)\\frac{63}{16}=ar^{3}\left(1+r+r^{2}\right)\end{array}\right.$$
- step2: Solve the equation:
  $$\left\{ \begin{array}{l}a=\frac{63}{2+2r+2r^{2}}\\frac{63}{16}=ar^{3}\left(1+r+r^{2}\right)\end{array}\right.$$
- step3: Substitute the value of $$a:$$
  $$\frac{63}{16}=\frac{63}{2+2r+2r^{2}}\times r^{3}\left(1+r+r^{2}\right)$$
- step4: Simplify:
  $$\frac{63}{16}=\frac{63r^{3}}{2}$$
- step5: Swap the sides:
  $$\frac{63r^{3}}{2}=\frac{63}{16}$$
- step6: Multiply both sides of the equation by $$2:$$
  $$\frac{63r^{3}}{2}\times 2=\frac{63}{16}\times 2$$
- step7: Multiply the terms:
  $$63r^{3}=\frac{63\times 2}{16}$$
- step8: Divide the terms:
  $$63r^{3}=\frac{63}{8}$$
- step9: Multiply by the reciprocal:
  $$63r^{3}\times \frac{1}{63}=\frac{63}{8}\times \frac{1}{63}$$
- step10: Multiply:
  $$r^{3}=\frac{1}{8}$$
- step11: Simplify the equation:
  $$\sqrt[3]{r^{3}}=\sqrt[3]{\frac{1}{8}}$$
- step12: Calculate:
  $$r=\sqrt[3]{\frac{1}{8}}$$
- step13: Simplify the root:
  $$r=\frac{1}{2}$$
- step14: Substitute the value of $$r:$$
  $$a=\frac{63}{2+2\times \frac{1}{2}+2\left(\frac{1}{2}\right)^{2}}$$
- step15: Substitute back:
  $$a=\frac{63}{2+1+2\times 2^{-2}}$$
- step16: Calculate:
  $$a=18$$
- step17: Calculate:
  $$\left\{ \begin{array}{l}a=18\r=\frac{1}{2}\end{array}\right.$$
- step18: Check the solution:
  $$\left\{ \begin{array}{l}a=18\r=\frac{1}{2}\end{array}\right.$$
- step19: Rewrite:
  $$\left(a,r\right) = \left(18,\frac{1}{2}\right)$$
The solution to the system of equations is $$ a = 18 $$ and $$ r = \frac{1}{2} $$ or $$ r = 0.5 $$.

Therefore, the value of $$ m $$ is $$ 18 $$ and the common ratio $$ r $$ is $$ \frac{1}{2} $$ or $$ 0.5 $$.
$$ m = 18 $$ and the common ratio $$ r = 0.5 $$.

Calculate the value of .
The sum of the first three terms of a geometric sequence is . The sum of the
fourth, fifth and sixth term of the same sequence is . Determine the value of
common ratio .

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