How many terms of the arthmetic progression 45,39,33,.... must be taken so that sum is 180?

Answer

To solve the problem of how many terms of the arithmetic progression (AP) 45, 39, 33, ... must be taken so that the sum is 180, we can follow these steps:

### Step 1: Identify the first term (A) and common difference (D)
The first term \( A \) of the AP is 45, and the common difference \( D \) can be calculated as:
\[ D = 39 - 45 = -6 \]

### Step 2: Write the formula for the sum of the first n terms (S_n)
The formula for the sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
We know that \( S_n = 180 \).

### Step 3: Substitute the known values into the formula
Substituting \( A = 45 \), \( D = -6 \), and \( S_n = 180 \) into the formula gives:
\[ 180 = \frac{n}{2} \times (2 \times 45 + (n - 1)(-6)) \]

### Step 4: Simplify the equation
This simplifies to:
\[ 180 = \frac{n}{2} \times (90 - 6n + 6) \]
\[ 180 = \frac{n}{2} \times (96 - 6n) \]

### Step 5: Multiply both sides by 2 to eliminate the fraction
\[ 360 = n(96 - 6n) \]

### Step 6: Rearrange the equation
Expanding the right side gives:
\[ 360 = 96n - 6n^2 \]
Rearranging this leads to:
\[ 6n^2 - 96n + 360 = 0 \]

### Step 7: Divide the entire equation by 6 to simplify
\[ n^2 - 16n + 60 = 0 \]

### Step 8: Factor the quadratic equation
We need to factor the quadratic:
\[ (n - 6)(n - 10) = 0 \]

### Step 9: Solve for n
Setting each factor equal to zero gives us:
1. \( n - 6 = 0 \) → \( n = 6 \)
2. \( n - 10 = 0 \) → \( n = 10 \)

### Conclusion
Thus, the number of terms that must be taken from the arithmetic progression so that the sum is 180 can be either 6 or 10.

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