In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A geometric series has common ratio r and first term a. Given r ≠ 1 and a ≠ 0 (a) prove that $$S_n = \frac{a(1-r^n)}{1-r}$$ (4) Given also that S_{10} is four times S_{5} (b) find the exact value of r. (4)

A-Level Edexcel Maths Pure Arc Length: In this question you must show a

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 2

Question 1

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A geometric series has common rat... show full transcript

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 2

Step 1

(a) prove that

96%

114 rated

Answer

To prove the formula for the sum of the first n terms of a geometric series, we start with the definition:

Let the series be:

$S_n = a + ar + ar^2 + \ldots + ar^{n-1}$

To find the sum, we can multiply the entire series by the common ratio r:

$rS_n = ar + ar^2 + ar^3 + \ldots + ar^n$

Now, we subtract this second equation from the first:

$S_n - rS_n = a - ar^n$

Factoring out S_n on the left side gives:

$S_n(1 - r) = a(1 - r^n)$

Now, divide both sides by (1 - r) (since r eq 1):

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

Step 2

(b) find the exact value of r.

99%

104 rated

Answer

We know from part (a) that:

$S_{10} = \frac{a(1 - r^{10})}{1 - r}$ $S_{5} = \frac{a(1 - r^{5})}{1 - r}$

Given that:

$S_{10} = 4S_{5}$

Substituting the expressions for S_{10} and S_{5} gives:

$\frac{a(1 - r^{10})}{1 - r} = 4 \cdot \frac{a(1 - r^{5})}{1 - r}$

Since a ≠ 0, we can cancel a and (1 - r):

$1 - r^{10} = 4(1 - r^{5})$

Expanding the right side:

$1 - r^{10} = 4 - 4r^{5}$

Rearranging terms:

$r^{10} - 4r^{5} + 3 = 0$

Letting x = r^5, we rewrite the equation as:

$x^2 - 4x + 3 = 0$

Factoring gives:

$(x - 1)(x - 3) = 0$

Thus, we find:

$x = 1 ext{ or } x = 3$

Substituting back for r gives:

If x = 1, then $r^5 = 1 ightarrow r = 1$ (not valid, since r ≠ 1). If x = 3, then $r^5 = 3 ightarrow r = 3^{1/5}$ .

Thus, the exact value of r is:

$r = 3^{1/5}$

In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 2

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 2

(a) prove that

(b) find the exact value of r.

Join the A-Level students using SimpleStudy...