How many terms of the series –12, – 9, – 6,... must be taken that the sum may be 54?

Options

A. 6
B. 9
C. 12
D. 24

Answer

To find how many terms of the series -12, -9, -6,... must be taken so that the sum equals 54, we can follow these steps:

### Step 1: Identify the series
The series given is -12, -9, -6, ... This is an arithmetic progression (AP) where:
- The first term (a) = -12
- The common difference (d) = -9 - (-12) = 3

### Step 2: Write the formula for the sum of the first n terms of an AP
The sum of the first n terms (S_n) of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]

### Step 3: Set up the equation for the sum
We need the sum to equal 54:
\[ S_n = 54 \]
Substituting the values of a and d into the formula:
\[ 54 = \frac{n}{2} \times (2(-12) + (n - 1)(3)) \]

### Step 4: Simplify the equation
Now simplify the equation:
\[ 54 = \frac{n}{2} \times (-24 + 3n - 3) \]
\[ 54 = \frac{n}{2} \times (3n - 27) \]
Multiply both sides by 2 to eliminate the fraction:
\[ 108 = n(3n - 27) \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 3n^2 - 27n - 108 = 0 \]

### Step 6: Simplify the quadratic equation
Divide the entire equation by 3:
\[ n^2 - 9n - 36 = 0 \]

### Step 7: Factor the quadratic equation
Now we can factor the quadratic:
\[ (n - 12)(n + 3) = 0 \]

### Step 8: Solve for n
Setting each factor to zero gives:
1. \( n - 12 = 0 \) → \( n = 12 \)
2. \( n + 3 = 0 \) → \( n = -3 \) (not a valid solution since n must be positive)

Thus, the only valid solution is:
\[ n = 12 \]

### Conclusion
Therefore, **12 terms** of the series must be taken for the sum to equal 54.

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