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vredina [299]
3 years ago
14

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

Mathematics
1 answer:
omeli [17]3 years ago
8 0

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

T_5 = 160

T_7 = 40

Required

T_6

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

\frac{T_7}{T_5} = \frac{40}{160}

Divide the numerator and the denominator of the fraction by 40

\frac{T_7}{T_5} = \frac{1}{4} ----- equation 1

Recall that the formula of a GP is

T_n = a r^{n-1}

Where n is the nth term

So,

T_7 = a r^{6}

T_5 = a r^{4}

Substitute the above expression in equation 1

\frac{T_7}{T_5} = \frac{1}{4}  becomes

\frac{ar^6}{ar^4} = \frac{1}{4}

r^2 = \frac{1}{4}

Square root both sides

r = \sqrt{\frac{1}{4}}

r = ±\frac{1}{2}

Next, is to solve for the first term;

Using T_5 = a r^{4}

By substituting 160 for T5 and ±\frac{1}{2} for r;

We get

160 = a \frac{1}{2}^{4}

160 = a \frac{1}{16}

Multiply through by 16

16 * 160 = a \frac{1}{16} * 16

16 * 160 = a

2560 = a

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

T_n = a r^{n-1}

Here, n = 6;

T_6 = a r^{6-1}

T_6 = a r^5

T_6 = 2560 r^5

r = ±\frac{1}{2}

So,

T_6 = 2560( \frac{1}{2}^5) or T_6 = 2560( \frac{-1}{2}^5)

T_6 = 2560( \frac{1}{32}) or T_6 = 2560( \frac{-1}{32})

T_6 = 80 or T_6 = -80

T_6 =±80

Hence, the 6th term is positive/negative 80

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