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Andreyy89
3 years ago
6

Express the terms of the following geometric sequence recursively.

Mathematics
1 answer:
BabaBlast [244]3 years ago
4 0

Answer:

The most correct option for the recursive expression of the geometric sequence is;

4. t₁ = 7 and tₙ = 2·tₙ₋₁, for n > 2

Step-by-step explanation:

The general form for the nth term of a geometric sequence, aₙ is given as follows;

aₙ = a₁·r⁽ⁿ⁻¹⁾

Where;

a₁ = The first term

r = The common ratio

n = The number of terms

The given geometric sequence is 7, 14, 28, 56, 112

The common ratio, r = 14/7 = 25/14 = 56/58 = 112/56 = 2

r = 2

Let, 't₁', represent the first term of the geometric sequence

Therefore, the nth term of the geometric sequence is presented as follows;

tₙ = t₁·r⁽ⁿ⁻¹⁾ = t₁·2⁽ⁿ⁻¹⁾

tₙ =  t₁·2⁽ⁿ⁻¹⁾ = 2·t₁2⁽ⁿ⁻²⁾ = 2·tₙ₋₁

∴ tₙ = 2·tₙ₋₁, for n ≥ 2

Therefore, we have;

t₁ = 7 and tₙ = 2·tₙ₋₁, for n ≥ 2.

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If the ratio is 6:1 in 3 years, then Murphy is 9 and Spelman is 69.

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<em>Alternate solution method</em>

I find it easiest to add 12 years to the total (each ages by 6 years), which will give a total age of 90 in 6 years. At that time, Murphy is 1/7 of the total of ages, so dividing that sum into parts with the appropriate ratio gives m'=90/7=12 6/7; s'=77 1/7. So, m=6 6/7; s=71 1/7.

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<em>Comment on the problem</em>

We think there is a "typo" because the ratio of 6:1 means the future total should be a multiple of 7. Of course, 78+6 =84 is a multiple of 7, but adding 6 to the total will occur in 3 years, not 6 years.

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