(how does n = (l-a/d)+1

Question

So I was introduced to sequences and series in my class and understood each and every formula but couldn't understand n = ((l-a)/d)+1 where l = last term , a = first term and d = common difference in an arithmetic progression.please be easy on me.

Answer 1

In an A.P. containing n terms, we can construct an expression for the last term in the sequence. Using your notation,

$$l = a + (n-1)d.$$

You can see why this is true by applying it to an actual sequence, say $2,5,8,11,\dots$

By rearranging the above equation, we then obtain $n = \frac{l-a}{d}+1$.

EDIT: Rearranging: $$ \begin{align} l-a &= (n-1)d \\ \frac{l-a}{d} &= n-1 \\ n &= \frac{l-a}{d}+1 \end{align}$$

Answer 2

Suppose that there are $n$ terms, starting with $a$ and having common difference $d$; then the terms are

$$a,a+d,a+2d,a+3d,\ldots,a+(n-1)d\;.$$

Why $n-1$ and not $n$? The $1$-st term in that list is $a+0d$, the $2$-nd term is $a+1d$, the $3$-rd term is $a+2d$, and in general the $k$-th term is $a+(k-1)d$. Thus, the $n$-th term is $a+(n-1)d$.

To put it a little differently, if we call the terms $a_1,a_2,\ldots,a_n$, to get from $a_1$ to $a_n$ we have to add $d$ once for each term from $a_2$ through $a_n$, and there are $n-1$ of them.

Thus, the last term must actually be $\ell=a+(n-1)d$, and therefore

$$\begin{align*} \frac{\ell-a}d+1&=\frac{\big(a+(n-1)d\big)-a}d+1\\ &=\frac{(n-1)d}d+1\\ &=(n-1)+1\\ &=n\;. \end{align*}$$

Answer 3

$$\underbrace{0\,\overbrace{\cdots\cdots0\cdots\cdots0\cdots\cdots0\cdots\cdots0\cdots\cdots0\cdots\cdots0\cdots\cdots0}^{l-a}}_{n\ {\rm terms}\ \ \therefore\ \ n-1\ {\rm gaps\ of\ size}\ d}$$

Answer 4

Let $S$ denote the sum of all the terms in the arithmetic progression.

Clearly,

\begin{align} S &= a + (a+d) + (a+2d) + \hspace{2mm}... \hspace{2mm}+ (a + (n-1)d) \hspace{40mm} [1] \\ S &= 0 + \hspace{2mm} a \hspace{9mm} + (a + \hspace{2mm}d) + \hspace{2mm} ... \hspace{2mm} + \hspace{1mm}(a + (n-2)d) + (a + (n-1)d) \hspace{6mm} [2]\\ \end{align}

Subtracting $[2]$ from $[1]$, we have: \begin{align} 0 &= a + d + d \hspace{1mm}+ \hspace{2mm} ... + \hspace{1mm} d \hspace{1mm} - (a + (n-1)d) \\ &= a + (n-1)d - l \end{align} Where $l$ denotes the last term in the arithmetic progression.

Hence, the equation now becomes: \begin{align} (n-1)*d &= (l-a) \\ (n-1) &= (l-a)/d \\ n &= (l-a)/d + 1 \end{align}