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A geometric series is a + ar + ar² + .. - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 3

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A geometric series is a + ar + ar² + ... (a) Prove that the sum of the first n terms of this series is given by $$S_n = \frac{a(1 - r^n)}{1 - r}$$ The third and f... show full transcript

Worked Solution & Example Answer:A geometric series is a + ar + ar² + .. - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 3

Step 1

Prove that the sum of the first n terms of this series is given by

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Answer

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

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

Multiplying both sides by the common ratio rr, we get:

rSn=ar+ar2+ar3++arnrS_n = ar + ar^2 + ar^3 + \ldots + ar^n

Subtracting these two equations:

SnrSn=aarnS_n - rS_n = a - ar^n

Factoring out SnS_n:

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

Rearranging gives:

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

Step 2

the common ratio

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Answer

Let the common ratio be rr. The third term is given by:

ar2=5.4ar^2 = 5.4

And the fifth term is:

ar4=1.944ar^4 = 1.944

Dividing the second equation by the first:

ar4ar2=1.9445.4\frac{ar^4}{ar^2} = \frac{1.944}{5.4}

This simplifies to:

r2=1.9445.4=0.36r^2 = \frac{1.944}{5.4} = 0.36

Taking the square root gives:

r=0.6r = 0.6.

Step 3

the first term

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Answer

Using the common ratio found, r=0.6r = 0.6. Substituting rr back into the equation from part (b):

ar2=5.4ar^2 = 5.4

Thus, we have:

a(0.6)2=5.4a(0.6)^2 = 5.4

Calculating gives:

0.36a=5.40.36a = 5.4

Hence,

a=5.40.36=15.a = \frac{5.4}{0.36} = 15.

Step 4

the sum to infinity

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Answer

The sum to infinity of a geometric series is given by the formula:

S=a1rS = \frac{a}{1 - r}

Substituting the values of aa and rr:

S=1510.6=150.4=37.5.S = \frac{15}{1 - 0.6} = \frac{15}{0.4} = 37.5.

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