Formula for the n-th term . Sequence Question

Hi, I am new to this forum and I am also very new to math. I failed math in high school so my understanding of algebra is very weak. Even so, I would like to learn so please bare with me if I am being a nuisance.

Question

Find formula, showing all the work for the n-th term tn of the sequence (tn) . Defined with

t1 = 8; tn = ( 5*tn-1 ): 2; *tn-1 = previous term

I understand a little bit of this

GCSE Maths Sequences Revision - The nth Term | S-cool, the revision website

however, I can't seems to relate this with the question I am facing.

Any help is appreciated

Thank you for your time

Re: Formula for the n-th term . Sequence Question

Each term is 5 times higher than the previous term, sequences like these are called geometric sequences, since the first term is 8 and the common ratio is 5 the formula for the nth term is 8*5^{n-1}

Re: Formula for the n-th term . Sequence Question

Thanks a lot dude! that explains a lot

Re: Formula for the n-th term . Sequence Question

Sorry actually I am a bit lost

If t1 = 8; and tn = (5*tn-1) /2

Then t3 = (5*20)/2 = 50

But using the n-th term mentioned wont it be

t(3) = 8*5^n-1

= 8*5^2

= 8*25

= 400

I am very new to this, so I'm sorry if i misunderstood you somewhere

Re: Formula for the n-th term . Sequence Question

Quote:

Originally Posted by

**RedGuitar** Sorry actually I am a bit lost

If t1 = 8; and tn = (5*tn-1) /2

Then t3 = (5*20)/2 = 50

But using the n-th term mentioned wont it be

t(3) = 8*5^n-1

= 8*5^2

= 8*25

= 400

I am very new to this, so I'm sorry if i misunderstood you somewhere

Your confusion results from your inability to post clearly what you mean.

Is it now $\displaystyle t_1=8~\&~t_{n}=\left(\frac{5}{2}\right)t_{n-1}~?$

1 Attachment(s)

Re: Formula for the n-th term . Sequence Question

The formula given is

Attachment 27480

Re: Formula for the n-th term . Sequence Question

Quote:

Originally Posted by

**RedGuitar**

Why don't you see that $\displaystyle \left( {\frac{{5{t_{n - 1}}}}{2}} \right) = \left( {\frac{5}{2}} \right){t_{n - 1}}~?$

Re: Formula for the n-th term . Sequence Question

Oh I see it now,

How does sequence is normally solved? Do we eye ball for a pattern or is there a magic formula for it?

Plato can you please show me step by step to find the n term, please? I've been stuck with this for hours : (.

I tried to follow this Geometric Sequences and Series but I am still confused.

Re: Formula for the n-th term . Sequence Question

Quote:

Originally Posted by

**RedGuitar** How does sequence is normally solved? Do we eye ball for a pattern or is there a magic formula for it?

$\displaystyle {t_n} = 8 \cdot {\left( {\frac{5}{2}} \right)^{n - 1}}$

Re: Formula for the n-th term . Sequence Question

Thanks a lot, I assume nth term = dn + (a - d) comes in play?

Thank you for your time everyone, it's been a great help

Re: Formula for the n-th term . Sequence Question

Quote:

Originally Posted by

**RedGuitar** Thanks a lot, I assume nth term = dn + (a - d) comes in play?

Thank you for your time everyone, it's been a great help

That is for an arithmetic sequence, this is a geometric sequence.