Find the number of terms in the arithmetic sequence $6 + 10 + 14 + \cdots + (4n-2)$

Question

I am trying to understand how to find the numbers of terms in the arithmetic sequence below.

$6 + 10 + 14 + ... + (4n-2)$

I think it should be $(n-1)$ because $$\frac{(4n-2)-6}{4} = \frac{4n-8}{4}=n-2$$

and we add a $1$ to $n-2$ since we undercount the first number (example, if we just had the sequence $6+10+14$ and we want to find the number of integers we can subtract $6$ from 14 and divide by $4$ i.e.,$\frac{14-6}{4}=2$, but we undercounted by $1$ so the total number of integers in the example sequence is $3$), so it should $n-1$ terms in the above arithmetic sequence, but the book says $n-2$ and I don't know why that is.

Answer 1

You are right; the book is wrong.

Your first term $6 = 4(2)-2$

Your final term is $4(n)-2$

The number of terms is the number of integers from $2$ to $n$ inclusive. Starting from $1$, that would be $n$, so here it's $n-1$.

Answer 2

The $n^{th}$ term of an arithmetic progression is given by $$= a +(n-1)d.$$
For your question,
$$4n-2= 6 + (y-1)4.$$
On solving this you get $$ y= n - 1.$$
Hence, the total number of terms is $$ n-1$$