Answer:

The 8th term of the Geometric Progression is -160.

Step-by-step explanation:

→LET US FIND FIRST THE RATIO TO DETERMINE THE SEQUENCE.

FORMULA:

\: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{an = a1 {r}^{n - 1}}}

an=a1r

n−1

GIVEN:

→ LET US SAY THAT THE 3RD TERM IS THE FIRST TERM AND THE 6TH TERM IS THE 4TH TERM.

\: \: \: \: \: \: \: \: a1 = 5 - \: the \: first \: term \: \:a1=5−thefirstterm

\: \: \: \: \: \: \: an = - 40 - \: the \: last \: terman=−40−thelastterm

\: \: \: \: \: \: \: \: \: n = 4 - \: the \: no. \: of \: termsn=4−theno.ofterms

SUBSTITUTE:

\: \: \: \: \: \: \: \: \: \: \: \: - 40 = 5 {r}^{4 - 1}−40=5r

4−1

SOLVE:

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{ - 40}{5} = \frac{5}{5} {r}^{4}

5

−40

=

5

5

r

4

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 8 = {r}^{4}−8=r

4

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sqrt[3]{ - 8} = \sqrt[3]{ {r}^{3} }

3

−8

=

3

r

3

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\boxed{- 2 = r}}

−2=r

GEOMETRIC SEQUENCE:

→ NOW THAT YOU KNOW THE RATIO IS -2, YOU CAN NOW DETERMINE THE GEOMETRIC PROGRESSION BY MULTIPLYING IT.

\: \:\green{\boxed{ \frac{5}{4}, \: - \frac{5}{2} , \: 5 , \: - 10 , \:20 ,\: \: - 40 , \: 80 ,\: \: - 160}}

4

5

,−

2

5

,5,−10,20,−40,80,−160

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