Question

28. The first and last terms of a geometric series are 2 and 2048 respectively. The sum of the series is
2730. Find the common ratio.
2 Points)​

Answer 1

Answer:

The formula for the nth term of a geometric sequence is:

a%5Bn%5D+=+a%5B1%5D%2Ar%5E%28%28n-1%29%29

with a%5B1%5D 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%2Ar%5E%28%28n-1%29%29

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.

If we don't know this about 2048 we could try to go further. Dividing both sides by 2 we get:

1024+=+r%5E%28%28n-1%29%29

Next we will use the formula for the sum of a geometric series:

S%5Bn%5D+=+a%5B1%5D%28%281-r%5En%29%2F%281-r%29%29

where S%5Bn%5D is the sum of the first n terms, a%5B1%5D 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%28%281-r%5En%29%2F%281-r%29%29

Multiplying each side by (1-r) [to eliminate the fraction] we get:

273%281-r%29+=+2%281-r%5En%29

which simplifies to:

273-273r+=+2%281-r%5En%29

Next we return to our earlier equation:

1024+=+r%5E%28%28n-1%29%29

If we multiply each side of this by r we get:

1024%2Ar+=+r%5E%28%28n-1%29%29+%2Ar

On the right side we use the rule for exponents when multiplying (i.e. add the exponents:

1024%2Ar+=+r%5E%28%28n-1%29%29+%2Ar%5E1

1024r+=+r%5E%28%28n-1%29%2B1%29

1024r+=+r%5En

This gives us an expression to use back in:

273-273r+=+2%281-r%5En%29

Substituting in 1024r:

273-273r+=+2%281-1024r%29

Now we can solve for r. Multiplying out the right side:

273-273r+=+2-2048r

Adding 2048r:

273%2B1775r+=+2

Subtracting 273:

1775r+=+-271

Dividing by 1775:

r+=+%28-271%29%2F1775

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!).

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.

Please mark me as brainliest

Thank you and have a nice day

Answer 2

Answer:

$\bigstar{\bold{Common\:ratio=4}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• First term of the G.P (a)= 2
• Last term of the G.P($ar^{n-1}$) = 2048
• Sum of the series ($S_n$) = 2730

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Common ratio (r)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ We have to find the common ratio of this geometric progression.

→ The sum of  terms of a G.P is given by the formula,

   $S_n=\dfrac{a(r^{n}-1) }{r-1}$

→ We know that $ar^{n-1}$ = 2048, a = 2

→ Hence,

  $2\times r^{n-1}=2048$

  $r^{n-1} =1024$

  $\dfrac{r^{n} }{r} =1024$

 $r^{n}=1024 r----(1)$

→ Substitute equation 1 and the other datas given in the formula above

  $2730=\dfrac{2\times( 1024r-1)}{r-1}$

→ Cross multiplying and simplifying,

   2730 (r - 1) = 2048 r - 2

   2730r - 2730 = 2048 r - 2

   2730r - 2048r = -2 + 2730

   682 r = 2728

           r = 2728/682

           r = 4

→ Hence thhe common ratio is 4

$\boxed{\bold{Common\:ratio=4}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ To find sum of n terms of a G.P,

→ If r > 1,

  $S_n=\dfrac{a(r^{n}-1) }{r-1}$

→ If r < 1

  $S_n=\dfrac{a(1-r^{n}) }{1-r}$