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The first three terms of a geometric series are 18, 12 and p respectively, where p is a constant - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 4

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The first three terms of a geometric series are 18, 12 and p respectively, where p is a constant. Find (a) the value of the common ratio of the series, (b) the va... show full transcript

Worked Solution & Example Answer:The first three terms of a geometric series are 18, 12 and p respectively, where p is a constant - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 4

Step 1

the value of the common ratio of the series

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Answer

To find the common ratio (r) of the geometric series, we use the formula:

r=a2a1=1218=23r = \frac{a_2}{a_1} = \frac{12}{18} = \frac{2}{3}

Therefore, the value of the common ratio is 23\frac{2}{3}.

Step 2

the value of p

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Answer

In a geometric series, the relationship between the terms can be represented as:

p=a2r=1223=8p = a_2 \cdot r = 12 \cdot \frac{2}{3} = 8

Thus, the value of p is 8.

Step 3

the sum of the first 15 terms of the series, giving your answer to 3 decimal places

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Answer

The formula for the sum of the first n terms of a geometric series is:

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

Where:

  • a = first term = 18
  • r = common ratio = 23\frac{2}{3}
  • n = number of terms = 15

Now, substituting the values into the formula:

S15=18(1(23)15)123S_{15} = \frac{18(1 - (\frac{2}{3})^{15})}{1 - \frac{2}{3}}

Calculating:

  • 1123=3\frac{1}{1 - \frac{2}{3}} = 3
  • The term (23)15\left(\frac{2}{3}\right)^{15} can be calculated to find its value.

Thus, S15=183(1(23)15)53.877S_{15} = 18 \cdot 3 \cdot \left(1 - (\frac{2}{3})^{15}\right)\approx 53.877

Therefore, the sum of the first 15 terms of the series is approximately 53.877.

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