Siva, a money lender, lends in the following manner. For the first year, he charges 2% p.a. for the second year, he charges 4% p.a. for the third year, the charges 6% p.a. and so on. If he lends a sum in this way at simple interest, find the least integral number of years in which it will fetch an interest at least equal to itself.

To find the least integral number of years in which the sum will fetch an interest at least equal to itself, we can start by setting up an equation.

Let's assume the sum of money lent is 'P' and the number of years is 'n'.

According to the given information, the interest charged for the first year is 2% of the sum, for the second year it is 4% of the sum, for the third year it is 6% of the sum, and so on.

So, the interest earned after 'n' years can be calculated using the formula:

Interest = P * (2 + 4 + 6 + ... + 2n)

This can be simplified as:

Interest = P * (2 + 4 + 6 + ... + 2n) = P * n * (n + 1)

Now, we need to find the least integral value of 'n' for which the interest earned is at least equal to the sum 'P'.

Setting up the equation:

P * n * (n + 1) >= P

Simplifying the equation:

n * (n + 1) >= 1

n^2 + n - 1 >= 0

Using the quadratic formula, we can find the roots of this equation:

n = (-1 ± √(1^2 - 4(1)(-1))) / 2

n = (-1 ± √(1 + 4)) / 2

n = (-1 ± √5) / 2

Since we are looking for the least integral value of 'n', we take the positive root:

n = (-1 + √5) / 2

Approximating this value, we get:

n ≈ 0.618

Since 'n' represents the number of years, we round up this value to the nearest whole number:

n = 1

Therefore, the least integral number of years in which the sum will fetch an interest at least equal to itself is 1 year.