How many terms of the A.P. 9, 17, 25, …..must be taken to give a sum of 636 ?

Answer

To solve the problem of finding how many terms of the arithmetic progression (A.P.) 9, 17, 25, ... must be taken to give a sum of 636, we will follow these steps:

### Step 1: Identify the first term and common difference
- The first term \( a \) of the A.P. is 9.
- The common difference \( d \) can be found by subtracting the first term from the second term:
\[ d = 17 - 9 = 8 \]

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

### Step 3: Substitute the known values into the formula
We know \( S_n = 636 \), \( a = 9 \), and \( d = 8 \). Substituting these values into the formula gives:
\[ 636 = \frac{n}{2} \times (2 \times 9 + (n - 1) \times 8) \]
This simplifies to:
\[ 636 = \frac{n}{2} \times (18 + (n - 1) \times 8) \]

### Step 4: Simplify the equation
Now, we can simplify the equation further:
\[ 636 = \frac{n}{2} \times (18 + 8n - 8) \]
\[ 636 = \frac{n}{2} \times (8n + 10) \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 1272 = n \times (8n + 10) \]
This expands to:
\[ 1272 = 8n^2 + 10n \]

### Step 5: Rearrange into standard quadratic form
Rearranging the equation gives:
\[ 8n^2 + 10n - 1272 = 0 \]

### Step 6: Simplify the quadratic equation
We can divide the entire equation by 2 to simplify it:
\[ 4n^2 + 5n - 636 = 0 \]

### Step 7: Factor the quadratic equation
Next, we will factor the quadratic equation. We look for two numbers that multiply to \( 4 \times -636 = -2544 \) and add to \( 5 \). The factors are \( 64 \) and \( -39 \):
\[ 4n^2 + 64n - 39n - 636 = 0 \]
Grouping gives:
\[ 4n(n + 16) - 39(n + 16) = 0 \]
Factoring out \( (n + 16) \):
\[ (n + 16)(4n - 39) = 0 \]

### Step 8: Solve for \( n \)
Setting each factor to zero gives:
1. \( n + 16 = 0 \) → \( n = -16 \) (not valid since \( n \) must be positive)
2. \( 4n - 39 = 0 \) → \( n = \frac{39}{4} = 9.75 \) (not valid since \( n \) must be a whole number)

### Step 9: Check for integer solutions
Since we need \( n \) to be a whole number, we can check the integer values around \( 9.75 \). Testing \( n = 12 \):
\[ S_{12} = \frac{12}{2} \times (18 + 8 \times 11) = 6 \times (18 + 88) = 6 \times 106 = 636 \]
Thus, the number of terms \( n \) that must be taken to give a sum of 636 is:
\[ \boxed{12} \]