What is 15/20 reduced
2 answers:
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For a geometric sequence:
= 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:
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:
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
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:
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:
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