Question 732891
THE LUCKY GUESS APPROACH:
If we call the number of terms {{{n}}}, and the common ratio {{{r}}}, the last term is
{{{b[n]=7*r^(n-1)=448}}} --> {{{r^(n-1)=448/7}}} --> {{{r^(n-1)=64}}} --> 
{{{r^(n-1)=2^6}}}
If we are lucky {{{r=2}}} and {{{n-1=6}}} --> {{{n=7}}}.
Those numbers would certainly agree with the first term being 7 and the last term being 448.
Would the sum of those 7 terms be 889?
We can figure out the sum using some formula from the book, or we can calculate terms number 2, 3, 4, 5, and 6 and then add all 7 terms.
Either way, we would find out that the sum of the first 7 terms of a geometric progression with ratio 2 and first term 7 is indeed 889.
The approach worked because the problem was design to have nice small numbers as answers for {{{r}}} and {{{n}}}.
 
ANOTHER APPROACH:
The sum of {{{n}}} terms, from {{{b[1]=7}}} to {{{b[n]=448}}} is {{{889}}}
The sum of the first {{{n-1}}} terms, from {{{b[1]=7}}} to {{{b[n-1]}}} is
{{{S=b[1]+b[2]}}}+ ... +{{{b[n-2]+b[n-1]=889-448=441}}}
If we call the common ratio {{{r}}},
{{{S*r=b[1]*r+b[2]*r}}}+... +{{{b[n-2]*r+b[n-1]*r=b[2]+b[3]}}}+ ... +{{{b[n-1]+b[n]}}}
because {{{b[1]*r=b[2]}}} , {{{b[2]*r=b[3]}}} , and so on until {{{b[n-2]*r=b[n-1]}}} , and {{{b[n-2]*r=b[n-1]}}} 
{{{S*r-S=b[n]-b[1]}}} --> {{{S*(r-1)=b[n]-b[1]}}} --> {{{r-1=(b[n]-b[1])/S}}} --> {{{r=1+(b[n]-b[1])/S}}}
Since we know {{{b[n]}}} , {{{b[1]}}}, and {{{S}}} we can calculate
{{{r=1+(448-7)/441}}} --> {{{r=1+441/441}}} --> {{{r=1+1}}} --> {{{highlight(r=2)}}}