Question

What is the value of a(1) for a geometric sequence with a(4) = 40 and a(6) = 160? A. 2 B. 5 C. 8 D. 10

Answer 1

well you need to find the common ratio 1st

\[160 = a_{1} \times r^{6 - 1}\]

and the 2nd equation is

\[40 = a_{1} \times r^{4 - 1}\]

you may be able to guess the common ratio....

but rewriting the 1st equation

\[160 = a_{1} \times r^3 \times r^2\]

substituting the 2nd equation

you get

\[160 = 40 \times r^2\]

solve for r...

then substitute it into either equation to find the 1st term

Answer 2

There is a way to work this out neatly, but I used to cheat on these questions and just input each answer into a calculator and fish for the correct answer.

a_6=a_1*r^(n-1)

160 = a * r^(6-1)

If you know r then you only need to solve for a

we know here that
40 * r * r = 160
so we can solve for r...

40 * r * r /40 = 160/40
r * r = 4
r^2 = 4
SQRT(r^2) = SQRT(4)
r = 2

so now we have
160 = a * 2^(5)

and now can you solve for a?

Answer 3

Thanks I got it!