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}
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
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
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.