Question

The sum of all natural number between 100 and 300 which when divided by 4 leaves remainder 3 is .

• A. 5025
• B. 5050
• C. 10050
• D. 10100

Answer 1

The correct option is C 10050

The first natural above 100 which leaves remainder 3 when divided by 4 is 103.
The next are 107, 111, 115,.... till 299.
Therefore, the sequence is 103, 107, 111, 115,...,299 which is an AP with first term 103 and common difference d=4.
Let there be n terms.
$a_n=299\Rightarrow a+(n-1)d=299\Rightarrow 103+(n-1)\times 4=299\Rightarrow (n-1)\times 4=196\Rightarrow n-1=49\Rightarrow n=50S_n=\frac{n}{2}\left(a_1+a_n\right)=\frac{50}{2}\left(103+299\right)S_n=10050$