Question

The last two digits in $X=\sum _{k=1}^{100}k$ are

• A. $50$
• B. $51$
• C. $52$
• D. $53$

Answer 1

The correct option is A $50$

We have to find the sum of 1+2+3+4....upto...100.

1,2,3,4 .....100

We see these numbers are in AP having common difference of 1.

So we apply here the formula :-

Sⁿ = n/2 ( a + l )

Here a = 1.

l = 100

n = 100

Sⁿ = 100/2 ( 1 + 100 )

➡ 100/2 ( 101 )

➡ 50 × 101

➡ 5050.

Required answer is :-

Sum of 1+2+3+4......100 is 5050.