Question

A new row is formed by the corresponding terms of a geometric sequence and an arithmetic sequence together.
The constant ratio of the geometric sequence 2 and the common difference of the arithmetic sequence 2.
The first term of the new row is 1, and the second term is 7, how would I go about calculating the third term of the new row? and writing an expression for the nth term of the new sequence

Answer 1

The sum of the corresponding terms?
So if Gn is the nth term of the Geometric sequence and An is the nth term of the Arithmatic sequence, then Cn (the nth term of the combined sequence) would be An + Gn?

Answer 2

Yes

Answer 3

An = A1 +2(n-1) and
Gn = G1*2^(n-1)
So Cn = An + Gn = (A1 + 2(n-1)) + (G1*2^(n-1))

Answer 4

2 equations and 2 unknowns:
term 1 gives us:
C1 = [A1 + 2(1-1)] + [G1*2^(1-1)] = A1 + G1 = 1
term 2 gives us:
C2 = [A1 + 2(2-1)] + [G1*2^(2-1)] = A1 + 2 + G2*2 = 7

Answer 5

A1 + G1 = 1
A1 + 2G1 = 5
so
A1 = 1-G1
(1-G1) + 2G1 = 5
so
G1 = 4 and
A1 = 1-4 = -3

Answer 6

Understand so far?

Answer 7

Ask me questions about what I did and why.

Answer 8

I follow you, Im trying hard since all my exams are bilingual but I've always learnt it in another language so I'm trying to figure out what your A1 and G1 values are which I suspect must be the arithmetic and geometric sequence... hmm I've got it so far yes

Answer 9

but how did you formulate the two equations?

Answer 10

A1 means the first term of the arithmatic sequence. G1 means the first terms of the geometric.

Answer 11

If you know the first term of an arithmatic sequence and the difference added each time, then you can write a formula for the nth term:
An = A1 +d(n-1)
where A1 is the first time and d is the common difference.

Answer 12

ahhh I see

Answer 13

Yes? And for the geometric sequence, I start with some initial value, and then I multiply by the same ratio every time.

Answer 14

ahhh ofcourse... yes because they said r = 2 and d = 2... dammit >.<

Answer 15

So the geometric sequence becomes:
Gn = G1 * r^(n-1)
where r is that common ratio. In this case it was 2

Answer 16

Haha yes. Do you understand those equations? The arithmetic one has you start with the initial value and then add the difference (n-1) times. The geometric one has you start with the initial value and multiply by the ratio (n-1) times.

Answer 17

I see... thank you very much, I get this now... I'm going to try the others myself if I struggle again can I post another question?

Answer 18

Yes, of course =D. Do you understand how to do the rest of the problem? We found the initial values for both. Now we use those initial values to write the formula for Cn, the nth term of the new sequence.

Answer 19

hmmm, dangit I forgot the rest no I'm not really sure, this is really new work to us and I got called out during class I'm a bit behind at the moment

Answer 20

Well, once you know the initial term of the arithmetic sequence, A1, and the difference, d, you can write the formula for An as
An = A1 + d(n-1)
Once you know the initial term of the geometric sequence, G1, and the ratio, r, you can write the formula for Gn as
Gn = G1*r^(n-1)

Then, since Cn comes from adding corresponding terms,
Cn = An + Gn

Answer 21

Okay yes it makes sence, awh super easy thank you very much :)

Answer 22

I've got it now, I'll be able to do these again... I hope... probably not hehe but stay positive I must thank you very much!!!! :D