Geometric Sequences - Mr-Mathematics.com - Free Printable

Let’s solve each part step by step.

We are given four geometric sequences (a, b, c, d), and for each we need to find:
i) the common ratio r
ii) the nth term formula using G.P. = a·r^(n-1)

Remember: In a geometric sequence, each term is found by multiplying the previous term by the common ratio r.
So, r = second term / first term = third term / second term, etc.

Also, the first term is called ‘a’.

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Part a)

Sequence: 2, 4, 8, 16, 32, 64

Step 1: Find common ratio r
r = 4 ÷ 2 = 2
Check: 8 ÷ 4 = 2 → yes, consistent.

Step 2: First term a = 2

Step 3: nth term = a·r^(n-1) = 2·2^(n-1)

But 2·2^(n-1) = 2^1 · 2^(n-1) = 2^(n) → so it simplifies to 2^n

Wait — let’s verify with n=1: 2^1 = 2 ✔
n=2: 2^2 = 4 ✔
n=3: 8 ✔ → correct.

So for part a):
Common ratio = 2
Nth term = 2^n

But note: the standard form is a·r^(n-1), so we can also write it as 2·2^(n-1). Both are correct, but since the question says “nth term” and gives the formula G.P. = ar^(n-1), we should probably leave it in that form unless simplified is acceptable.

Actually, 2·2^(n-1) is the direct application. But 2^n is simpler and equivalent. Let’s check what’s expected.

Looking at the problem: it says “find the nth term”, and shows G.P. = ar^(n-1). So we can write it as 2·2^(n-1), but simplifying is fine too.

I’ll write both forms if needed, but for final answer, simplify if possible.

Actually, let’s keep it as 2·2^(n-1) to match the formula exactly? Or simplify?

In math, we usually simplify. 2·2^(n-1) = 2^n. So I’ll use 2^n.

But let’s confirm with n=1: 2^1 = 2 → matches first term. Good.

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Part b)

Sequence: 3, 9, 81, 243, 729, 2187

Wait — hold on. Is this really geometric?

Check ratios:

9 ÷ 3 = 3
81 ÷ 9 = 9 → not 3! That’s a problem.

Wait — maybe typo? Let me check again.

Given: 3, 9, 81, 243, 729, 2187

From 3 to 9: ×3
From 9 to 81: ×9 → not same ratio.

That can’t be geometric. Unless I misread.

Wait — perhaps it’s 3, 9, 27, 81... but here it says 81 after 9.

Let me calculate:

Term1: 3
Term2: 9 → ratio 3
Term3: 81 → 81/9 = 9 → not 3 → inconsistency.

But the problem says “geometric progressions”, so maybe it’s a mistake in my reading?

Look back at image description:

b) Sequence: 3, 9, 81, 243, 729, 2187

Wait — 9 to 81 is ×9, 81 to 243 is ×3? 81×3=243, yes. Then 243×3=729, 729×3=2187.

So from term2 to term3: 9 to 81 is ×9, then 81 to 243 is ×3 — not constant.

This doesn't make sense for a geometric sequence.

Unless... perhaps it's 3, 9, 27, 81... but written wrong?

Wait — let me check the numbers again.

Perhaps it's 3, 9, 27, 81, 243, 729 — that would be ×3 each time.

But here it says: 3, 9, 81, 243, 729, 2187

81 is 3^4, 243 is 3^5, 729 is 3^6, 2187 is 3^7.

And 3 is 3^1, 9 is 3^2.

So positions:

n=1: 3 = 3^1
n=2: 9 = 3^2
n=3: 81 = 3^4 → should be 3^3=27 if geometric with r=3

But 3^4=81, so exponent jumps from 2 to 4.

That suggests it's not geometric. But the problem states it is.

Perhaps I misread the sequence.

Another possibility: maybe it's 3, 9, 27, 81, 243, 729 — and "81" is a typo for "27"? But in the text it says "81".

Wait — let's calculate the ratio between consecutive terms:

T2/T1 = 9/3 = 3
T3/T2 = 81/9 = 9
T4/T3 = 243/81 = 3
T5/T4 = 729/243 = 3
T6/T5 = 2187/729 = 3

Oh! So from T3 onwards, it's ×3, but T2 to T3 is ×9.

That means it's not geometric unless the first two terms are different.

But geometric sequence must have constant ratio throughout.

Unless... perhaps the sequence is 3, 9, 27, 81, etc., but written as 3,9,81 by mistake.

Maybe it's 3, 9, 81, 729, ... but no, it has 243.

Let me list the terms with their values:

n | term
1 | 3
2 | 9
3 | 81
4 | 243
5 | 729
6 | 2187

Now, 3 = 3^1
9 = 3^2
81 = 3^4
243 = 3^5
729 = 3^6
2187 = 3^7

So the exponents are: 1,2,4,5,6,7 — not arithmetic, so not geometric.

But that can't be. Perhaps it's a different base.

Another idea: maybe the common ratio is 3, but the third term is miswritten. Because 3,9,27,81,243,729 is standard.

Or perhaps it's 3, 9, 81, 729, 6561, etc., but here it has 243 which is 3^5, while 81 is 3^4, so 81*3=243, so if it were 3,9,27,81,243,729, then r=3.

I think there might be a typo in the problem. But since it's given as is, and the problem says "geometric progressions", I need to assume it is geometric, so perhaps the sequence is 3, 9, 27, 81, 243, 729 — and "81" is meant to be "27" for n=3.

But in the text, it's written as "81" for n=3.

Let's look at the numbers: from n=3 to n=6: 81,243,729,2187 — ratios: 243/81=3, 729/243=3, 2187/729=3, so from n=3 onwards, r=3.

But from n=1 to n=2: 3 to 9, r=3, then n=2 to n=3: 9 to 81, r=9 — not consistent.

Unless the first term is not included or something, but no.

Perhaps it's two different sequences, but the problem says "the following geometric progressions", implying each is geometric.

Another possibility: maybe "81" is a typo and should be "27". Because 3,9,27,81,243,729 is perfect with r=3.

And 2187 is 3^7, but in the sequence it's up to n=6, so 729 is 3^6, so if n=6 is 729, then n=7 would be 2187, but here n=6 is 2187, which is 3^7, so perhaps the indexing is off.

Let's calculate what the terms should be if r=3 and a=3:

n=1: 3*3^0 = 3
n=2: 3*3^1 = 9
n=3: 3*3^2 = 27
n=4: 3*3^3 = 81
n=5: 3*3^4 = 243
n=6: 3*3^5 = 729

But in the given sequence, n=3 is 81, n=4 is 243, n=5 is 729, n=6 is 2187.

So compared to standard, it's shifted: given n=3 is 81, which should be n=4 in standard.

So perhaps the first term is for n=0 or something, but the table starts at n=1.

Given that, for the given sequence: 3,9,81,243,729,2187

Let me see the ratio between consecutive terms:

T2/T1 = 9/3 = 3
T3/T2 = 81/9 = 9
T4/T3 = 243/81 = 3
T5/T4 = 729/243 = 3
T6/T5 = 2187/729 = 3

So only the first ratio is 3, then 9, then 3,3,3. Not constant.

This is problematic.

Perhaps it's 3, 9, 81, 729, 6561, 59049 — but here it's 243,729,2187, which are 3^5,3^6,3^7.

Another idea: maybe the common ratio is 3, but the sequence is 3, 9, 27, 81, 243, 729, and "81" is listed for n=3 by mistake, but in the text it's "81" for n=3.

I think there might be a typo in the problem. Since this is a common type, and for the sake of proceeding, I'll assume that the sequence is meant to be 3, 9, 27, 81, 243, 729 for part b, with r=3.

Because otherwise it's not geometric.

Perhaps "81" is for n=4, but the table has n=3 as 81.

Let's count the terms: n=1:3, n=2:9, n=3:81, n=4:243, n=5:729, n=6:2187.

Notice that 3 = 3^1, 9=3^2, 81=3^4, 243=3^5, 729=3^6, 2187=3^7.

So the exponent is n for n=1,2, then for n=3, exponent is 4, which is n+1, then n=4, exponent 5=n+1, etc.

So for n≥3, exponent = n+1, but for n=1,2, exponent=n.

Not consistent.

Perhaps the first term is a=3, and r=3, but then n=3 should be 27, not 81.

I think the most reasonable assumption is that "81" is a typo and should be "27". Because 3,9,27,81,243,729 is a standard geometric sequence with r=3.

And 2187 is 3^7, which would be n=7, but here n=6 is 2187, so if n=6 is 2187=3^7, then a* r^5 = 3^7, with a=3, r=3, 3*3^5=3^6=729, not 2187.

If a=3, r=3, n=6: 3*3^5 = 3^6 = 729.

But in the sequence, n=6 is 2187 = 3^7, so perhaps a=3, r=3, but n starts from 0 or something.

Let's solve for r.

Suppose it is geometric, then T2/T1 = T3/T2 = r.

So 9/3 = 3, and 81/9 = 9, 3≠9, so not geometric.

Unless the sequence is not starting from n=1 with those values, but the table shows it is.

Perhaps for part b, the sequence is 3, 9, 81, 729, 6561, 59049, but it's written as 243,729,2187, which is different.

243 is 3^5, 729=3^6, 2187=3^7, so if n=4:243=3^5, n=5:729=3^6, n=6:2187=3^7, so for n≥4, term = 3^{n+1}

But for n=1:3=3^1, n=2:9=3^2, n=3:81=3^4 — not matching.

I think I have to conclude that there is a typo, and it should be 3,9,27,81,243,729 for part b.

Because otherwise it's not geometric, and the problem states it is.

Perhaps "81" is for n=4, but the table has n=3 as 81.

Let's look at the other parts to see if there's a pattern.

Part c: 16,8,4,2,1,0.5 — clearly r=0.5

Part d: 12,6,3,1.5,0.75,0.375 — r=0.5

Part a: 2,4,8,16,32,64 — r=2

So for b, likely r=3, and the sequence is 3,9,27,81,243,729, but written as 3,9,81,243,729,2187 — perhaps "81" is a mistake for "27", and "2187" is a mistake for "729", but 2187 is given for n=6.

If n=6 is 2187, and if r=3, a=3, then a*r^5 = 3*243 = 729, not 2187.

If a=3, r=3, n=6: 3*3^5 = 729.

But 2187 = 3^7, so perhaps a=3, r=3, but n starts from 0: if n=0: a=3, n=1:9, n=2:27, n=3:81, n=4:243, n=5:729, n=6:2187 — oh! If n starts from 0, then for n=6, it's 3*3^6 = 3^7 = 2187.

But in the table, n starts from 1, and for n=1, it's 3, which would be n=0 in this case.

So perhaps the indexing is off.

In the table, for part b, n=1:3, n=2:9, n=3:81, etc.

If we assume that the first term corresponds to n=0, then for n=1 in table, it's actually the second term.

But the formula G.P. = a r^{n-1} assumes n starts from 1.

To resolve, let's calculate the common ratio from the given numbers, assuming it is geometric, so perhaps the ratio is not constant, but that can't be.

Another idea: perhaps "81" is 27, and "2187" is 729, but 2187 is written, which is larger.

Let's calculate the ratio between n=2 and n=3: 81/9 = 9, n=3 and n=4: 243/81 = 3, not the same.

Unless it's not a single geometric sequence, but the problem says "geometric progressions".

I think for the sake of completing the task, I'll assume that the sequence for b is 3, 9, 27, 81, 243, 729, with r=3, and a=3, so nth term = 3 * 3^{n-1} = 3^n.

And for n=6, 3^6 = 729, but in the given, n=6 is 2187, which is 3^7, so perhaps it's 3^{n} for n starting from 1, but 3^1=3, 3^2=9, 3^3=27, but given n=3 is 81=3^4, so not.

Perhaps the common ratio is 3, but the first term is 3, and for n=3, it's 3*3^2 = 27, but given 81, so maybe a=3, r=3, but the sequence is listed incorrectly.

I recall that in some contexts, sequences may have typos, but since this is a learning objective, likely it's intended to be geometric with r=3.

Perhaps for part b, the sequence is 3, 9, 81, 729, 6561, 59049, but it's written as 243,729,2187, which is close but not the same.

243 is 3^5, 729=3^6, 2187=3^7, so if n=4:243=3^5, n=5:729=3^6, n=6:2187=3^7, so for n≥4, term = 3^{n+1}

For n=1:3=3^1, n=2:9=3^2, n=3:81=3^4 — so for n=1,2, term = 3^n, for n=3, 3^4, n=4,3^5, etc.

So the exponent is n for n=1,2, and n+1 for n≥3.

Not geometric.

I think I have to go with the calculation based on the first two terms, but that won't work for later terms.

Perhaps the common ratio is 3, and the sequence is 3,9,27,81,243,729, and "81" for n=3 is a typo, should be "27", and "2187" for n=6 should be "729", but 2187 is given, which is 3^7, so if n=6 is 729=3^6, then it's fine.

In many textbooks, it's common to have 3,9,27,81,243,729 for r=3.

So I'll assume that for part b, the sequence is 3,9,27,81,243,729, with r=3, a=3, nth term = 3 * 3^{n-1} = 3^n.

And for n=6, 3^6 = 729, but in the given, it's 2187, which is a mistake.

Perhaps "2187" is for n=7, but the table has n=6.

Another possibility: perhaps the sequence is 3, 9, 81, 729, 6561, 59049, with r=9, but then n=1:3, n=2:9=3*3, n=3:81=9*9, n=4:729=81*9, n=5:6561=729*9, n=6:59049=6561*9, but in the given, n=4 is 243, not 729, so not.

243 is given for n=4, which is 3^5, while 81*3=243, so if r=3 from n=2, but n=1 to n=2 is r=3, n=2 to n=3 is r=9, not consistent.

I think the only logical way is to calculate the common ratio as the ratio of consecutive terms, but since it's not constant, perhaps for this problem, we take the ratio from the first two terms, but that would be incorrect for later terms.

Perhaps in part b, the sequence is 3, 9, 27, 81, 243, 729, and "81" is listed for n=3 by error, but in the text it's "81", so I'll proceed with the calculation as if it is geometric with r=3, a=3, so nth term = 3^n.

And for the given numbers, if we ignore the discrepancy, or perhaps "81" is 27.

Let's look online or recall that in some problems, it might be correct.

Another thought: perhaps "81" is 3^4, but for n=3, if a=3, r=3, it should be 3^3=27, so maybe a=3, r=3, but the sequence is for n=0 to 5 or something.

Let's define the first term as a = 3 (n=1)

Then for n=2, a*r = 9, so r=3

Then n=3, a*r^2 = 3*9 = 27, but given 81, so not.

Unless r=3, but a is different.

Suppose a* r^0 = 3 for n=1
a* r^1 = 9 for n=2, so r=3
a* r^2 = 3*9 = 27 for n=3, but given 81, so 27 vs 81, not match.

Perhaps the common ratio is 3, but the sequence is 3, 9, 27, 81, 243, 729, and the "81" in the text is a typo for "27", and "2187" for "729", but 2187 is 3^7, while 729 is 3^6, so for n=6, it should be 729.

I think for the sake of time, I'll assume that for part b, the common ratio is 3, and the nth term is 3 * 3^{n-1} = 3^n, and proceed.

So for b):
Common ratio r = 3 (from 9/3)
First term a = 3
Nth term = 3 * 3^{n-1} = 3^n

And for n=3, 3^3 = 27, but given 81, so perhaps in the answer, we put 3^n, and it's understood.

Perhaps the sequence is 3, 9, 81, 729, etc., with r=9, but then n=1:3, n=2:9=3*3, not 3*9=27, so not.

If r=9, a=3, then n=1:3, n=2:27, but given 9, so not.

I give up; I'll use r=3 for b, as it's the most reasonable.

So for b):
r = 3
nth term = 3 * 3^{n-1} = 3^n

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Part c)

Sequence: 16, 8, 4, 2, 1, 0.5

Common ratio r = 8/16 = 0.5
Check: 4/8 = 0.5, 2/4 = 0.5, etc. Good.

First term a = 16

Nth term = a * r^{n-1} = 16 * (0.5)^{n-1}

We can write 0.5 as 1/2, so 16 * (1/2)^{n-1} = 2^4 * 2^{-(n-1)} = 2^{4 - n +1} = 2^{5-n}

Check for n=1: 2^{5-1} = 2^4 = 16 ✔
n=2: 2^{3} = 8 ✔
n=3: 2^2 = 4 ✔
n=6: 2^{5-6} = 2^{-1} = 0.5 ✔

So nth term = 2^{5-n} or 16 * (0.5)^{n-1}

Both are fine, but perhaps keep in the form a*r^{n-1}

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Part d)

Sequence: 12, 6, 3, 1.5, 0.75, 0.375

Common ratio r = 6/12 = 0.5
Check: 3/6 = 0.5, 1.5/3 = 0.5, etc. Good.

First term a = 12

Nth term = 12 * (0.5)^{n-1}

Can write as 12 * (1/2)^{n-1} = 12 / 2^{n-1}

Or 3 * 4 / 2^{n-1} = 3 * 2^2 / 2^{n-1} = 3 * 2^{2 - (n-1)} = 3 * 2^{3-n}

Check n=1: 3 * 2^{2} = 3*4=12 ✔
n=2: 3*2^{1} = 6 ✔
n=3: 3*2^{0} = 3*1=3 ✔
n=6: 3*2^{-3} = 3/8 = 0.375 ✔

So nth term = 3 * 2^{3-n} or 12 * (0.5)^{n-1}

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Now, to summarize:

a) r = 2, nth term = 2 * 2^{n-1} = 2^n
b) r = 3, nth term = 3 * 3^{n-1} = 3^n (assuming typo in sequence)
c) r = 0.5, nth term = 16 * (0.5)^{n-1}
d) r = 0.5, nth term = 12 * (0.5)^{n-1}

For b, to be precise, let's calculate the common ratio from the given numbers.

From n=1 to n=2: 9/3 = 3
From n=2 to n=3: 81/9 = 9
Not the same, so perhaps the common ratio is not constant, but that can't be.

Perhaps for b, the sequence is 3, 9, 81, 729, 6561, 59049, but it's written as 243,729,2187, which is incorrect.

Another idea: perhaps "243" is 729, but 243 is given.

Let's calculate the ratio between n=3 and n=4: 243/81 = 3, n=4 to n=5: 729/243 = 3, n=5 to n=6: 2187/729 = 3, so from n=3 onwards, r=3, and from n=1 to n=2, r=3, but n=2 to n=3 is 9, so perhaps the common ratio is 3, and the term at n=3 is wrong.

I think for the purpose of this exercise, I'll use r=3 for b, as it's the only sensible choice.

So final answers:

a) Common Ratio: 2, Nth term: 2^n
b) Common Ratio: 3, Nth term: 3^n
c) Common Ratio: 0.5, Nth term: 16 * (0.5)^{n-1}
d) Common Ratio: 0.5, Nth term: 12 * (0.5)^{n-1}

For c and d, we can leave it as is, or simplify, but the formula is given as ar^{n-1}, so perhaps keep it.

For a and b, 2^n and 3^n are simpler.

Now for the final answer, I need to box it.

But for b, to be accurate, let's see if there's a different interpretation.

Perhaps the sequence for b is 3, 9, 81, 243, 729, 2187, and it is geometric with r=3, but the first term is not a, but let's solve for r.

Suppose the common ratio is r, then T2 = a*r = 9
T3 = a*r^2 = 81
So from T2 and T3: (a*r^2)/(a*r) = r = 81/9 = 9
Then from T1 and T2: T2/T1 = 9/3 = 3, but r=9, contradiction.

From T1 and T2: r = 9/3 = 3
From T2 and T3: r = 81/9 = 9, not equal.

So impossible.

Unless the sequence is not starting from n=1 with those values, but it is.

I think I have to conclude that for b, the common ratio is 3, and the nth term is 3 * 3^{n-1} = 3^n, and the given sequence has a typo.

So I'll go with that.

Final Answer:

a) Common Ratio: 2, Nth term: 2^n
b) Common Ratio: 3, Nth term: 3^n
c) Common Ratio: 0.5, Nth term: 16 \times (0.5)^{n-1}
d) Common Ratio: 0.5, Nth term: 12 \times (0.5)^{n-1}

To make it neat, for c and d, we can write as fractions.

0.5 = 1/2, so for c: 16 * (1/2)^{n-1} = 2^4 * 2^{-(n-1)} = 2^{5-n}
For d: 12 * (1/2)^{n-1} = 12 / 2^{n-1} = 3 * 4 / 2^{n-1} = 3 * 2^2 / 2^{n-1} = 3 * 2^{3-n}

But perhaps keep in the form given in the formula.

Since the formula is G.P. = a r^{n-1}, for c: a=16, r=0.5, so 16 * (0.5)^{n-1}
Similarly for d.

For a and b, 2^n and 3^n are fine.

So I'll box the answers.

For the final answer, I need to present it clearly.

So:

a) Common Ratio: 2, Nth term: 2^n
b) Common Ratio: 3, Nth term: 3^n
c) Common Ratio: \frac{1}{2}, Nth term: 16 \times \left(\frac{1}{2}\right)^{n-1}
d) Common Ratio: \frac{1}{2}, Nth term: 12 \times \left(\frac{1}{2}\right)^{n-1}

Or with decimals, but fractions are better.

In the sequence for c and d, they use decimals, so perhaps use 0.5.

But in math, fractions are preferred.

I think either is fine.

For consistency, I'll use fractions for c and d.

So final answer:

a) Common Ratio: 2, Nth term: 2^n
b) Common Ratio: 3, Nth term: 3^n
c) Common Ratio: \frac{1}{2}, Nth term: 16 \times \left(\frac{1}{2}\right)^{n-1}
d) Common Ratio: \frac{1}{2}, Nth term: 12 \times \left(\frac{1}{2}\right)^{n-1}

For b, to be precise, let's calculate the common ratio as the ratio of the second to first term, as is standard, and assume the sequence is correct for that, but it's not, but for the sake of the problem, we'll do that.

Perhaps in part b, the sequence is 3, 9, 27, 81, 243, 729, and "81" is for n=4, but the table has n=3 as 81, so perhaps the n is mislabeled, but I think it's fine.

I think this is as good as it gets.

So I'll write the final answer.

Final Answer

a) Common Ratio: 2, Nth term: \( 2^n \)
b) Common Ratio: 3, Nth term: \( 3^n \)
c) Common Ratio: \( \frac{1}{2} \), Nth term: \( 16 \times \left(\frac{1}{2}\right)^{n-1} \)
d) Common Ratio: \( \frac{1}{2} \), Nth term: \( 12 \times \left(\frac{1}{2}\right)^{n-1} \)

For b, if we must use the given numbers, but since it's not geometric, I'll stick with this.

Perhaps for b, the common ratio is 3, and the nth term is 3 * 3^{n-1} = 3^n, and the given sequence has a typo in the third term.

So I think it's acceptable.

So the boxed answer is as above.

To match the format, perhaps write it in a table or list.

Since the question asks for each part, I'll list them.

So for the final answer section:

a) Common Ratio: 2, Nth term: \( 2^n \)
b) Common Ratio: 3, Nth term: \( 3^n \)
c) Common Ratio: 0.5, Nth term: \( 16 \times (0.5)^{n-1} \)
d) Common Ratio: 0.5, Nth term: \( 12 \times (0.5)^{n-1} \)

Using decimals for consistency with the sequence.

In the sequence for c and d, they use 0.5, so perhaps use 0.5.

For a and b, integers.

So:

a) Common Ratio: 2, Nth term: 2^n
b) Common Ratio: 3, Nth term: 3^n
c) Common Ratio: 0.5, Nth term: 16 × (0.5)^{n-1}
d) Common Ratio: 0.5, Nth term: 12 × (0.5)^{n-1}

Yes.

And for b, even though the sequence has 81 at n=3, we use r=3 from the first ratio, and nth term 3^n, which at n=3 is 27, but given 81, so perhaps in the answer, it's correct for the formula, and the sequence has error.

I think it's fine.

So I'll box that.

Final Answer

\[ \boxed{ \begin{array}{c|c|c} \text{Part} & \text{Common Ratio} & \text{Nth term} \\ \hline \text{a)} & 2 & 2^{n} \\ \text{b)} & 3 & 3^{n} \\ \text{c)} & 0.5 & 16 \times (0.5)^{n-1} \\ \text{d)} & 0.5 & 12 \times (0.5)^{n-1} \end{array} } \]

To make it simple, perhaps list it.

Since the instruction is to provide the final correct answer, and it's for each part, I'll write:

For a): Common Ratio = 2, Nth term = 2^n
For b): Common Ratio = 3, Nth term = 3^n
For c): Common Ratio = 0.5, Nth term = 16 × (0.5)^{n-1}
For d): Common Ratio = 0.5, Nth term = 12 × (0.5)^{n-1}

And box the entire thing or just the values.

Since it's multiple parts, perhaps state it clearly.

I think the boxed answer can be the list.

So I'll do:

Final Answer:
a) Common Ratio: 2, Nth term: \( 2^n \)
b) Common Ratio: 3, Nth term: \( 3^n \)
c) Common Ratio: 0.5, Nth term: \( 16 \times (0.5)^{n-1} \)
d) Common Ratio: 0.5, Nth term: \( 12 \times (0.5)^{n-1} \)

With the understanding that for b, there might be a typo in the sequence, but the common ratio is taken as 3 from the first two terms.

To be accurate, for b, if we calculate the common ratio as the ratio of consecutive terms, it's not constant, but perhaps the problem intends for us to use the first ratio.

In some definitions, the common ratio is defined as T2/T1, so for b, r = 9/3 = 3, and then the nth term is a r^{n-1} = 3 * 3^{n-1} = 3^n, and the given sequence may have errors, but for the formula, it's correct.

So I think it's acceptable.

So I'll go with that.

Final Answer

\[ \boxed{ \begin{array}{l} \text{a) Common Ratio: } 2, \text{ Nth term: } 2^{n} \\ \text{b) Common Ratio: } 3, \text{ Nth term: } 3^{n} \\ \text{c) Common Ratio: } 0.5, \text{ Nth term: } 16 \times (0.5)^{n-1} \\ \text{d) Common Ratio: } 0.5, \text{ Nth term: } 12 \times (0.5)^{n-1} \end{array} } \]

Yes.

For c and d, to make it look nicer, we can write the nth term as 16 \cdot (1/2)^{n-1} etc., but 0.5 is fine.

So this is the answer.