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Find the geometric sequence given the sum of all the terms is 3339, the last term is 1696, and the common ratio is two.
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In order to answer this question, we will need to recall two formulae.
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First, π sub π or the πth term is equal to π multiplied by π to the power of π minus one, where π is the first term and π is the common ratio of the geometric sequence.
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The second formula tells us that the sum of the first π terms, written π sub π, is equal to π multiplied by one minus π to the power of π or divided by one minus π.
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When the common ratio is greater than one, as in this case, we tend to rewrite this as π multiplied by π to the power of π minus one all divided by π minus one.
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We are told that the last term is equal to 1696.
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Using the fact that π is equal to two, this gives us π multiplied by two to the power of π minus one is equal to 1696.
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Our laws of indices or exponents mean that we can rewrite two to the power of π minus one as two to the power of π divided by two to the power of one.
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We can then multiply both sides of this equation by two such that π multiplied by two to the power of π is equal to 3392.
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As we have two unknowns in this equation, we will stop here and call this equation one.
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We are also told in the question that the sum of all the terms is 3339.
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Distributing the parentheses or expanding the brackets of our formula along with substituting π is equal to two gives us π multiplied by two to the power of π minus π all divided by two minus one is equal to 3339.
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As two minus one is equal to one, we have π multiplied by two to the power of π minus π is equal to 3339.
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We will call this equation two.
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We can now substitute our value of π multiplied by two to the power of π into equation two.
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3392 minus π is equal to 3339.
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We can add π and subtract 3339 from both sides of this equation.
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This gives us a value of π equal to 53.
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The first term of the geometric sequence is 53.
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The geometric sequence contains the terms 53, 106, 212, and so on all the way up to 1696.