What is the greatest possible sum of the A.P. 17,14,11,…

Options

A. 51
B. 73
C. 57
D. 75

Answer

To find the greatest possible sum of the arithmetic progression (A.P.) given by the terms 17, 14, 11, ..., we can follow these steps:

### Step 1: Identify the first term and common difference
The first term \( a \) of the A.P. is 17, and the common difference \( d \) can be calculated as:
\[ d = 14 - 17 = -3 \]

### Step 2: Write the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting the values of \( a \) and \( d \):
\[ S_n = \frac{n}{2} \times (2 \times 17 + (n - 1)(-3)) \]

### Step 3: Simplify the sum formula
Now, simplify the equation:
\[ S_n = \frac{n}{2} \times (34 - 3n + 3) = \frac{n}{2} \times (37 - 3n) \]
\[ S_n = \frac{n(37 - 3n)}{2} \]

### Step 4: Differentiate to find the maximum sum
To maximize \( S_n \), we differentiate it with respect to \( n \):
\[ \frac{dS_n}{dn} = \frac{1}{2}(37 - 6n) \]
Setting the derivative equal to zero to find critical points:
\[ 37 - 6n = 0 \implies n = \frac{37}{6} \]

### Step 5: Verify that this is a maximum
To confirm that this critical point gives a maximum, we can perform the second derivative test:
\[ \frac{d^2S_n}{dn^2} = -3 \]
Since this is less than zero, it indicates that \( S_n \) has a maximum at \( n = \frac{37}{6} \).

### Step 6: Calculate the maximum sum
Now substitute \( n = \frac{37}{6} \) back into the sum formula:
\[ S_n = \frac{\frac{37}{6}(37 - 3 \times \frac{37}{6})}{2} \]
Calculating \( 37 - 3 \times \frac{37}{6} \):
\[ = 37 - \frac{111}{6} = \frac{222 - 111}{6} = \frac{111}{6} \]
Thus,
\[ S_n = \frac{\frac{37}{6} \times \frac{111}{6}}{2} = \frac{37 \times 111}{72} \]
Calculating \( 37 \times 111 = 4097 \):
\[ S_n = \frac{4097}{72} \]

### Step 7: Final calculation
Now, we can simplify \( \frac{4097}{72} \) to find the approximate value:
\[ \approx 57 \]

### Conclusion
The greatest possible sum of the A.P. is approximately **57**.