If the first term `a_(1)` of an A.P. is 22, the common difference `d=-4` and the sum to n terms is 64, find n. Explain the double answer.

Answer

To solve the problem step by step, we will use the formula for the sum of the first n terms of an arithmetic progression (A.P.).

### Given:
- First term \( a_1 = 22 \)
- Common difference \( d = -4 \)
- Sum of the first n terms \( S_n = 64 \)

### Step 1: Write down the formula for the sum of the first n terms of an A.P.
The formula for the sum of the first n terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

### Step 2: Substitute the known values into the formula.
Substituting \( S_n = 64 \), \( a = 22 \), and \( d = -4 \) into the formula:
\[ 64 = \frac{n}{2} \left( 2 \cdot 22 + (n - 1)(-4) \right) \]
This simplifies to:
\[ 64 = \frac{n}{2} \left( 44 - 4(n - 1) \right) \]

### Step 3: Simplify the equation.
Now, simplify the expression inside the parentheses:
\[ 44 - 4(n - 1) = 44 - 4n + 4 = 48 - 4n \]
So, we have:
\[ 64 = \frac{n}{2} (48 - 4n) \]

### Step 4: Multiply both sides by 2 to eliminate the fraction.
\[ 128 = n(48 - 4n) \]

### Step 5: Expand and rearrange the equation.
Expanding the right side gives:
\[ 128 = 48n - 4n^2 \]
Rearranging this leads to:
\[ 4n^2 - 48n + 128 = 0 \]

### Step 6: Divide the entire equation by 4 to simplify.
\[ n^2 - 12n + 32 = 0 \]

### Step 7: Factor the quadratic equation.
To factor \( n^2 - 12n + 32 = 0 \):
\[ (n - 4)(n - 8) = 0 \]

### Step 8: Solve for n.
Setting each factor to zero gives:
\[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \]
\[ n - 8 = 0 \quad \Rightarrow \quad n = 8 \]

### Step 9: Explain the double answer.
We have two possible values for \( n \): 4 and 8.

- If \( n = 4 \), the sum of the first four terms (22, 18, 14, 10) is:
\[ 22 + 18 + 14 + 10 = 64 \]
- If \( n = 8 \), the sum of the first eight terms includes the first four terms (which sum to 64) and the next four terms (which are negative and sum to 0). Thus, the total remains 64.

### Conclusion:
The two values of \( n \) indicate that the sum can be achieved with either the first four terms or the first eight terms of the sequence, where the latter includes terms that cancel each other out.

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