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In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 2 - 2019 - Paper 1

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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 2 - 2019 - Paper 1

Step 1

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

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Answer

To prove the formula for the sum of the first nn terms of a geometric series, we start by writing out the series terms:

Sn=a+ar+ar2+...+arn1S_n = a + ar + ar^2 + ... + ar^{n-1}

Next, we can express this sum in terms of the common ratio:

Sn=a(1+r+r2+...+rn1)S_n = a(1 + r + r^2 + ... + r^{n-1})

Now, we multiply both sides of the equation by (1r)(1 - r):

(1r)Sn=(1r)(a+ar+ar2+...+arn1)(1 - r)S_n = (1 - r)(a + ar + ar^2 + ... + ar^{n-1})

This simplifies to:

(1r)Sn=aarn(1 - r)S_n = a - ar^n

By distributing (1r)(1 - r), we can rearrange:

SnrSn=aarnS_n - rS_n = a - ar^n

Combining terms gives:

Sn(1r)=a(1rn)S_n(1 - r) = a(1 - r^n)

Finally, we divide both sides by (1r)(1 - r) (noting that r1r \neq 1) to arrive at the desired result:

Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r}

Step 2

find the exact value of $r$

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Answer

From the problem, we know that:

S10=4SsS_{10} = 4S_s

Using the formula derived in part (a), we write:

S10=a(1r10)1rS_{10} = \frac{a(1 - r^{10})}{1 - r}

and

Ss=a(1rs)1rS_s = \frac{a(1 - r^s)}{1 - r}

Substituting these into the equation gives:

a(1r10)1r=4a(1rs)1r\frac{a(1 - r^{10})}{1 - r} = 4 \frac{a(1 - r^s)}{1 - r}

As a0a \neq 0, we can cancel aa from both sides:

1r101r=41rs1r\frac{1 - r^{10}}{1 - r} = 4 \frac{1 - r^s}{1 - r}

This simplifies to:

1r10=4(1rs)1 - r^{10} = 4(1 - r^s)

Rearranging gives:

1r10=44rs1 - r^{10} = 4 - 4r^s

Thus, we get:

4rs=3r104r^s = 3 - r^{10}

To find the value of rr, we would need the value of ss. Assuming ss is defined per specific conditions or values, we can solve for rr accordingly using additional equations or iterating possible values of rr within the range of geometric series.

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