For a geometric sequence:
[tex] U_{n} = a * r^{n - 1} [/tex]
[tex] U_{n} [/tex] = the nth term, i.e. U₂ is the second term, U₃ is the third term, etc.
a = first term (U₁)
r = multiplier
n = number corresponding to the postion of the nth term in the sequence, e.g. for U₂, n = 2, for U₃, n = 3
According to the information:
U₂ = 6
U₄ = 54
So:
[tex] U_{2} = a * r^{(2) - 1} = 6 \\\\ a * r^{1} = 6 \\\\ ar = 6 \\[/tex]
[tex] U_{4} = a * r^{(4) - 1} = 54 \\\\ a * r^{3} = 54 \\\\ ar^{3} = 54[/tex]
Now, we have two equations in terms of a and r;
We can work out a by first working out r;
If we let the U₂ eq'n be [1] and the U₄ eq'n be [2];
Then, we do [2] ÷ [1] to cancel out the a and get an equation with only r that we are able to solve and then just solve it, like so:
[tex] \frac{ar^{3}}{ar} = \frac{54}{6} \\\\ r^{2} = 9 \\\\ r = 3 [/tex]
Note: r has to be 3, and not -3, as the two given terms of the sequence tell us that the terms of the sequence will get bigger from each term to the next so r has to be above 1.
Now that we know r, we can sub it back into either [1] or [2] and solve for a, Its quicker and easier to sub it into [1] as we have no powers greater than 1 and we have smaller numbers to deal with (but you would get the same value of a either way), so:
U₂ = ar = 6
a(3) = 6
a = 6/3
a = 2
