1 + 4 + 7 + 10 + … + x = 590, then the value of x is

Options

A. 55
B. 58
C. 61
D. None of these

Answer

To solve the equation \( 1 + 4 + 7 + 10 + \ldots + x = 590 \), we first recognize that the series is an arithmetic progression (AP) where the first term \( a = 1 \) and the common difference \( d = 3 \).

### Step 1: Identify the number of terms
The general formula for the \( n \)-th term of an AP is given by:
\[ a_n = a + (n-1)d \]
In this case, we can express \( x \) (the last term) as:
\[ x = 1 + (n-1) \cdot 3 \]
This simplifies to:
\[ x = 3n - 2 \]

### Step 2: Use the formula for the sum of the first \( n \) terms
The sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} (2a + (n-1)d) \]
Substituting the known values:
\[ 590 = \frac{n}{2} (2 \cdot 1 + (n-1) \cdot 3) \]
This simplifies to:
\[ 590 = \frac{n}{2} (2 + 3n - 3) \]
\[ 590 = \frac{n}{2} (3n - 1) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 1180 = n(3n - 1) \]
Rearranging gives:
\[ 3n^2 - n - 1180 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -1 \), and \( c = -1180 \):
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 3 \cdot (-1180) \]
Calculating this:
\[ 1 + 14160 = 14161 \]
Now, taking the square root:
\[ \sqrt{14161} = 119 \]
Substituting back into the quadratic formula:
\[ n = \frac{1 \pm 119}{6} \]
Calculating the two possible values for \( n \):
1. \( n = \frac{120}{6} = 20 \)
2. \( n = \frac{-118}{6} \) (not valid since \( n \) must be positive)

Thus, \( n = 20 \).

### Step 4: Find the value of \( x \)
Now we substitute \( n \) back into the equation for \( x \):
\[ x = 3n - 2 = 3 \cdot 20 - 2 = 60 - 2 = 58 \]

### Final Answer
The value of \( x \) is \( \boxed{58} \).