Question

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?

A) positive/negative 70
B) 70
C) positive/negative 80
D) 80

Answer 1

C. The 6th term is positive/negative 80

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

`(T_7)/(T_5) = (40)/(160)`

Divide the numerator and the denominator of the fraction by 40

`(T_7)/(T_5) = (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

`(T_7)/(T_5) = (1)/(4)` becomes

`(ar^6)/(ar^4) = (1)/(4)`

`r^2 = (1)/(4)`

Square root both sides

`r = \sqrt{(1)/(4)}`

r = ±
`(1)/(2)`

Next, is to solve for the first term;

Using
`T_5 = a r^(4)`

By substituting 160 for T5 and ±
`(1)/(2)` for r;

We get

`160 = a (1)/(2)^(4)`

`160 = a (1)/(16)`

Multiply through by 16

`16 * 160 = a (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 = ±
`(1)/(2)`

So,

`T_6 = 2560( (1)/(2)^5)` or
`T_6 = 2560( (-1)/(2)^5)`

`T_6 = 2560( (1)/(32))` or
`T_6 = 2560( (-1)/(32))`

`T_6 = 80` or
`T_6 = -80`

`T_6 =`±80

Hence, the 6th term is positive/negative 80