Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000. Class 9

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000. Class 9

Question 1.

Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000. Class 9

Solution:

The sum of the first n natural numbers is

S_n = $\frac{n(n+1)}{2}$

To find the smallest natural number n such that, $\frac{n(n+1)}{2}$ > 1000.

Multiplying 2 on both sides, we get

n(n + 1) > 2000 ...(i)

To estimate the value of n, look for a perfect square near 2000.

Since 40² = 1600 and 50² = 2500,

∴ Trying values in the mid-40s:

When n = 44, 44 × 45 =1980 < 2000

When n = 45, 45 × 46 = 2070 > 2000.

∴ smallest value of n for which inequality (i) is satisfied is n = 45.

∴ S₄₅ = $\frac{45 \times 46}{2}$ = 1035

∴ The sum of the first 45 natural number is greater than 1000.

∴ Smallest value of n is 45.