Geometric progression has 6 terms. If the 3rd and 4th term are 28-and 56 find the sum of the geometric progression
Let the first term of the geometric progression be $a$ and the common ratio be $r$. So we have the following terms:
\[a, ar, ar^2, ar^3, ar^4, ar^5.\]
We are given that the 3rd term is 28, so $ar^2=28$. Similarly, the 4th term is 56, so $ar^3=56$. Dividing these two equations, we get $\frac{ar^2}{ar^3}=\frac{28}{56}=\frac{1}{2}$. Canceling the $a$ term, we have $\frac{r^2}{r^3}=\frac{1}{2}$. Multiplying both sides by $r$ gives $r=\frac{1}{2}$.
Substituting $r=\frac{1}{2}$ into $ar^2=28$, we get $a\left(\frac{1}{2}\right)^2=28$. This simplifies to $a\cdot\frac{1}{4}=28$, or $a=28\cdot 4=112$.
The sum of a geometric series is $\frac{a(1-r^n)}{1-r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. Substituting in our values, we get $\frac{112\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}=\frac{112\left(1-\frac{1}{64}\right)}{\frac{1}{2}}$. Simplifying further, we find $\frac{112\cdot\frac{63}{64}}{\frac{1}{2}}=\frac{112\cdot 63\cdot 2}{64}=\frac{112\cdot 63}{32}$ $=7\cdot 63=\boxed{441}$.