How many terms of the series 18 + 15 + 12 +…………... when added together will give 45 ?

Answer

To solve the problem of how many terms of the series 18 + 15 + 12 + … when added together will give 45, we can follow these steps:

### Step 1: Identify the first term and common difference
The given series is an arithmetic progression (AP).
- The first term \( a = 18 \)
- The second term \( 15 \)
- The third term \( 12 \)

To find the common difference \( d \):
\[ d = 15 - 18 = -3 \]
or
\[ d = 12 - 15 = -3 \]

### Step 2: Set up the formula for the sum of the first \( n \) terms
The sum of the first \( n \) terms of an arithmetic progression is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]

We need to find \( n \) such that \( S_n = 45 \).

### Step 3: Substitute the known values into the formula
Substituting \( S_n = 45 \), \( a = 18 \), and \( d = -3 \) into the formula:
\[ 45 = \frac{n}{2} \times (2 \times 18 + (n - 1)(-3)) \]

### Step 4: Simplify the equation
Calculating \( 2a \):
\[ 2a = 2 \times 18 = 36 \]

Now substitute this back into the equation:
\[ 45 = \frac{n}{2} \times (36 + (n - 1)(-3)) \]
\[ 45 = \frac{n}{2} \times (36 - 3n + 3) \]
\[ 45 = \frac{n}{2} \times (39 - 3n) \]

### Step 5: Clear the fraction
Multiply both sides by 2 to eliminate the fraction:
\[ 90 = n(39 - 3n) \]
\[ 90 = 39n - 3n^2 \]

### Step 6: Rearrange into standard quadratic form
Rearranging gives:
\[ 3n^2 - 39n + 90 = 0 \]

### Step 7: Simplify the equation
Divide the entire equation by 3:
\[ n^2 - 13n + 30 = 0 \]

### Step 8: Factor the quadratic equation
Now we need to factor the quadratic:
\[ n^2 - 10n - 3n + 30 = 0 \]
\[ (n - 10)(n - 3) = 0 \]

### Step 9: Solve for \( n \)
Setting each factor to zero gives:
1. \( n - 10 = 0 \) → \( n = 10 \)
2. \( n - 3 = 0 \) → \( n = 3 \)

### Step 10: Conclusion
Thus, the number of terms required to be added to get a sum of 45 can be either 10 or 3.