Question

Find the values of t_1 and are for a geometric sequence with t_6= 4 and t_10 = 324.​

Answer 1

Answer: Choice C

$t_1 = \frac{4}{243}, \ r = 3$

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Work Shown:

Start with $t_6 = 4$ and multiply by the common ratio r to get the next term $t_7$. Keep this going until you get to the 10th term $t_{10}$

So we have...

$t_6 = 4\\\\t_7= r*t_6 = 4r\\\\t_8 = r*t_7 = r*(r*t_6) = r^2*t_6 = 4r^2\\\\t_9 = r*t_8 = r*(r^2*t_6) = r^3*t_6 = 4r^3\\\\t_{10} = r*t_9 = r*(r^3*t_6) = r^4*t_6 = 4r^4 = 324$

From that we can see,

$4r^4 = 324\\\\r^4 = \frac{324}{4}\\\\r^4 = 81\\\\r = 81^{1/4}\\\\r = 3$

At this point, the only possible answer is choice C since this is the only answer choice with r = 3 in it. We could stop here if we wanted.

For the sake of completeness, let's find $t_1$

$t_n = t_1*(r)^{n-1}\\\\t_{6} = t_1*(3)^{6-1}\\\\4 = t_1*(3)^{5}\\\\4 = t_1*243\\\\t_1 = \frac{4}{243}$

You can use $t_{10} = 324$ in place of $t_6 = 4$ in the section above, so either value is optional.