Find all possible ways of expressing 100 as the sum of consecutive natural numbers. Class 9

Find all possible ways of expressing 100 as the sum of consecutive natural numbers. Class 9

Question 1.

Find all possible ways of expressing 100 as the sum of consecutive natural numbers. Class 9

Solution:

Let the sum of consecutive numbers = 100, first term = a and number of terms = n.

Since S_n = $\frac{n}{2}$[a + l]

[result from Aryabhata's Aryabhatiya]

∴ S_n = $\frac{n}{2}$[a + (a + (n -1)(1))] = 100

[d = 1 since numbers are consecutive]

∴ n(2a + n - 1) = 200 ... (i)

∴ 2a + n - 1 = $\frac{200}{n}$

Now we try the values of n such that n divides 200 and 'a' comes out to be a positive integer.

Devisors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200 If we take n = 5,

If we take n = 5,

$\frac{200}{5}$ = 2a + 4 ... [From (i)]

∴ 40 = 2a + 4

36 = 2a

∴ a = 18

∴ The sum, 18 + 19 + 20 + 21 + 22 = 100

If we take n = 8,

$\frac{200}{8}$ = 2a + 7 ... [From (i)]

∴ 25 = 2a+ 7

18 = 2a

∴ a = 9

∴ The sum,

9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100

All other divisors of 200 give non-integer or negative values of a, hence rejected

∴ There are only 2 ways to express 100 as a sum of consecutive natural numbers i.e.,

100 = 18 + 19 + 20 + 21 + 22 and

100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.