CAT 2002 — QA Question 33

Sum of first n natural numbers = S(n)

Sum given by student = 575

S(10) = $\frac{10\times 11}{2}=$ 55

S(20) = $\frac{20\times 21}{2}=$ 210

S(30) = $\frac{30\times 31}{2}=$ 465

S(40) = $\frac{40\times 41}{2}=$ 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S(35) = $\frac{36\times 35}{2}=$ 630

The student stopped somewhere before 35.

∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595

But the student gave 575 as the sum, so the student missed on the number 20.

Hence, option 4.