Solve : 1 + 6 + 11 +….+ x = 148.

To solve the equation 1 + 6 + 11 + ... + x = 148, we need to find the value of x that satisfies the equation.

We notice that the given series follows an arithmetic progression with a common difference of 5. The first term (a) is 1, and the last term (x) is unknown. Let's use the formula for the sum of an arithmetic series to solve for x.

The sum of an arithmetic series is given by the formula:
S = (n/2) * (a + l),
where S is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case, the sum S is 148, the common difference d is 5, and the first term a is 1. We want to find the value of x, which is the last term.

Using the formula, we have:
148 = (n/2) * (1 + x).

To solve for x, we need to determine the number of terms (n) in the series. We can rearrange the equation to solve for n:

2 * 148 = n * (1 + x),
296 = n + nx.

Now, we can try different values of n and solve for x to find a solution that satisfies the equation.

Let's check for n = 1:
296 = 1 + 1x,
295 = x.
However, this does not satisfy the given series because the first term is 1 and the last term is 295, resulting in a large difference.

Let's check for n = 2:
296 = 2 + 2x,
294 = 2x,
x = 147.

Now, let's check for n = 3:
296 = 3 + 3x,
293 = 3x,
x = 97.67.

Since x should be a whole number to fit the arithmetic series, we can conclude that there is no integer solution for x that satisfies the equation 1 + 6 + 11 + ... + x = 148.