Question 825369
This problem is not possible.<br>
Here's why. The formula for the nth term of a geometric sequence is:
{{{a[n] = a[1]*r^((n-1))}}}
with {{{a[1]}}} being the first term, "r" being the common ratio, and "n" being the number of the term. So 2 and 2048 should fit this formula:
{{{2048 = 2*r^((n-1))}}}
If we realize that 2048 = 2*2*2*2*2*2*2*2*2*2*2 then we know that the only factors of 2048 are 2's or powers of 2. 2's and powers of 2 cannot add up to an odd number like 273.<br>
If we don't know this about 2048 we could try to go further. Dividing both sides by 2 we get:
{{{1024 = r^((n-1))}}}
Next we will use the formula for the sum of a geometric series:
{{{S[n] = a[1]((1-r^n)/(1-r))}}}
where {{{S[n]}}} is the sum of the first n terms, {{{a[1]}}} is the first term, "r" is the common ratio, and "n" is the number of terms. Substituting in our values for the sum and the first term we get:
{{{273 = 2((1-r^n)/(1-r))}}}
Multiplying each side by (1-r) [to eliminate the fraction] we get:
{{{273(1-r) = 2(1-r^n)}}}
which simplifies to:
{{{273-273r = 2(1-r^n)}}}
Next we return to our earlier equation:
{{{1024 = r^((n-1))}}}
If we multiply each side of this by r we get:
{{{1024*r = r^((n-1)) *r}}}
On the right side we use the rule for exponents when multiplying (i.e. add the exponents:
{{{1024*r = r^((n-1)) *r^1}}}
{{{1024r = r^((n-1)+1)}}}
{{{1024r = r^n}}}
This gives us an expression to use back in:
{{{273-273r = 2(1-r^n)}}}
Substituting in 1024r:
{{{273-273r = 2(1-1024r)}}}
Now we can solve for r. Multiplying out the right side:
{{{273-273r = 2-2048r}}}
Adding 2048r:
{{{273+1775r = 2}}}
Subtracting 273:
{{{1775r = -271}}}
Dividing by 1775:
{{{r = (-271)/1775}}}
With this value for "r", it will be impossible for the series to start at 2 and eventually get to 2048. (Try it and see!).<br>
Since the formulas for geometric series do not work with a first term of 2, a last term of 2048 and a sum of 273, these numbers cannot reflect a geometric series.