Fundamental Lines

$\text{The sum of the numbers between 1 and 81 divisible by 4 can be written}$ $\text{in the form of a AP like this :}$

$4,8,12.......76,80$

$\text{We can find the number of terms using the following formula,}$

$a_{n} = a+(n-1)\times d$

$\text{Where } a_{n} = 80 , a = 4 \text{ and } d = 4$

$\text{We get,}$

$n=20$

$\text{Then we can apply the formula}$

$S_{n} = \frac{n}{2}\times (a +l)$

$\text{Where } n = 20 ,a = 4 \text{ and } l = 80$

$\text{Now we can sub in the values and get } S_{20}$

$S_{20} = \boxed{840}$

4+8+...+80 is an AP with d=4. no.of terms in the series =((80-4)/4)+1=19+1=20 Sum of the series=20(4+80)/2=840

In this case it is just adding the multiples of 4 That is 4,8,12,16......,80 because the question says till 81

=> 4+8+12+.........+80 Take 4 common we get 4(1+2+3+4+......+20) =4 * 20(20+1)/2. Since(1+2+3+......+n)=n(n+1)/2 =4 (20)(21)/2 =4 (10)(21) = 840