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The fourth term of a geometric sequence is 48 - HSC - SSCE Mathematics Advanced - Question 21 - 2023 - Paper 1

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The fourth term of a geometric sequence is 48. The eighth term of the same sequence is \( \frac{3}{16} \). Find the possible value(s) of the common ratio and the c... show full transcript

Worked Solution & Example Answer:The fourth term of a geometric sequence is 48 - HSC - SSCE Mathematics Advanced - Question 21 - 2023 - Paper 1

Step 1

Let \( a \) = first term and \( r \) = common ratio

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Answer

From the geometric sequence, we can express:

  1. The fourth term: ( ar^3 = 48 )
  2. The eighth term: ( ar^7 = \frac{3}{16} )

These give us two equations to work with.

Step 2

Divide equation (2) by equation (1)

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Answer

Dividing ( ar^7 ) by ( ar^3 ), we obtain:

ar7ar3=31648\frac{ar^7}{ar^3} = \frac{\frac{3}{16}}{48}

This simplifies to:

r4=1256r^4 = \frac{1}{256}

Step 3

Solve for \( r \)

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Answer

Taking the fourth root of both sides gives:

r=14 or r=−14r = \frac{1}{4} \text{ or } r = -\frac{1}{4}

Step 4

Find corresponding first term(s)

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Answer

Substituting ( r = \frac{1}{4} ) into ( ar^3 = 48 ):

a(14)3=48a \left( \frac{1}{4} \right)^3 = 48
Thus, ( a \cdot \frac{1}{64} = 48 )
=> ( a = 48 \cdot 64 = 3072 )

For ( r = -\frac{1}{4} ):

a(−14)3=48a \left( -\frac{1}{4} \right)^3 = 48
So, ( a \cdot -\frac{1}{64} = 48 )
=> ( a = 48 \cdot -64 = -3072 $$

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