Algebra 2 - Geometric

Question

Find the 10th term -3/4, -3/2, -3.

 

I need to show work and I don't understand how to do this.

Answer 1

A geometric sequence has the general form:

 

a_n = a₁·r^n-1

• n = number of the term
• a_n = the nth term in the sequence
• a₁ = the first term in the sequence = -3/4
• r = the common ratio

 

T^o find the common ratio, r, take the ratio of any term with its preceding term:

 

r = (-3/2)/(-3/4) = (3/2)·(4/3) = 2

r = -3/(-3/2) = 3·(2/3) = 2

 

so the common ratio is r = 2 and the equation of the sequence is:

 

a_n = a₁·r^n-1

a_n = (-3/4)·2^n-1

 

To find the 10th term (a₁₀), set n=10 and use your calculator to compute the answer.

Answer 2

a1=-3/4,

a2=-3/2,

a3=-3,

....

q=a3/a2=(-3/(-3/2)=2.

a10=q^9*a1=2^9(-3/4)=-3*2^7=-384

 

In general,

a_n=q^n-1a_n-1