Q.Find the sum of all multiples of 5 between 1 and 100 (inclusive).

These multiples form an AP with $a=5, d=5$; find $n$ (using the last term $100$) then apply the sum formula.

$S_{20}=\dfrac{20}{2}(5+100)=10(105)=1050$.

The multiples of $5$ from $5$ to $100$ form an AP: $5,10,15,\ldots,100$, with $a=5$, $d=5$, and last term $l=100$.

First find how many terms there are: $l=a+(n-1)d \Rightarrow 100=5+(n-1)(5) \Rightarrow (n-1)=19 \Rightarrow n=20$.

Apply $S_n=\dfrac{n}{2}(a+l)$: $S_{20}=\dfrac{20}{2}(5+100)=10(105)=1050$.

Check by an independent method: every multiple of $5$ from $5$ to $100$ can be written as $5\times1, 5\times2,\ldots,5\times20$, so their sum is $5\times(1+2+\cdots+20)=5\times\dfrac{20(21)}{2}=5\times210=1050$ — using §6's $\sum i$ formula, matching exactly.

Alternative — sum formula in the other form: $S_{20}=\dfrac{20}{2}[2(5)+19(5)]=10[10+95]=10(105)=1050$, matching both other methods.

A common slip is including or excluding an endpoint incorrectly — since the range is "between 1 and 100, inclusive," $100$ itself (a multiple of 5) must be counted as the last term; excluding it would wrongly use $l=95$ and $n=19$.