Question

Find the sum of all natural numbers from 1 to 200 which are divisible by 4.

Answer 1

Hint: First find the natural numbers from 1 to 200 which are divisible by 4. The resulting sequence will be in A.P therefore we can use the sum of n terms of A.P to find the answer.

Complete step-by-step answer:
We basically need to find the arithmetic progression(A.P) from \[4+8+12+.....+200\]
So we do not know how many terms we have from 4 to 200.
Thus to find the number of terms we will solve the equation;
\[a+(n-1)d=200\]

Since we have the first term \[a=4\] and difference between two terms which is \[d=4\] . we get \[d=4\] because in the multiple of 4 every product has a difference of 4.
\[d=(12-8)\text{ or }d=(8-4)\]
Then we say that there are \[n\] terms and 200 is the nth term.
Next we solve the equation...
\[\begin{align}
  & d=(12-8)\text{ or }d=(8-4) \\
 & 4+(n-1)4=200 \\
 & (n-1)4=196 \\
 & (n-1)=49 \\
 & n=50 \\
\end{align}\]
 We get \[n=50\]
 Now, we put the value of \[n=50\] in the sum of the series equation of Arithmetic progression and we get...
\[\begin{align}
  & S=\dfrac{n}{2}\left( 2a+(n-1)d \right) \\
 & {{S}_{50}}=\dfrac{50}{2}\left( 2(4)+(50-1)4 \right) \\
 & {{S}_{50}}=25(8+(49\times 4)) \\
 & {{S}_{50}}=25\left( 8+196 \right)=25\times (204)=5100 \\
\end{align}\]
Therefore, the sum of all the natural numbers between 1 to 200 divisible by 4 is 5100.

Note: We can also find the number of terms using the concept of permutation and combination. But this is a more conceptual and easier method since we had the first term last term and the difference between two terms. To find the sum of series we can use an alternative formula which includes first and last term.