The 12 numbers, `a _(1), a _(2)………, a _(12)` are in arithmetical progression. The sum of al these numbers is 354. Let `P = a _(2) + a _(4) + ……………a _(12) and Q = a _(1) + a _(3) + ……..+a _(11) .` If the ratio P :Q is `32:27,` the common difference of the progression is

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the problem
We have 12 numbers in an arithmetic progression (AP) denoted as \( a_1, a_2, \ldots, a_{12} \). The sum of these numbers is given as 354.

### Step 2: Use the formula for the sum of an AP
The sum \( S_n \) of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For our case, \( n = 12 \), so:
\[ S_{12} = \frac{12}{2} \times (2a + 11d) = 6(2a + 11d) \]
Setting this equal to 354, we have:
\[ 6(2a + 11d) = 354 \]
Dividing both sides by 6:
\[ 2a + 11d = 59 \quad \text{(Equation 1)} \]

### Step 3: Define \( P \) and \( Q \)
Next, we define:
- \( P = a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} \)
- \( Q = a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} \)

### Step 4: Express \( P \) and \( Q \) in terms of \( a \) and \( d \)
The even indexed terms can be expressed as:
\[ P = (a + d) + (a + 3d) + (a + 5d) + (a + 7d) + (a + 9d) + (a + 11d) \]
This simplifies to:
\[ P = 6a + (1 + 3 + 5 + 7 + 9 + 11)d = 6a + 36d \]

The odd indexed terms can be expressed as:
\[ Q = a + (a + 2d) + (a + 4d) + (a + 6d) + (a + 8d) + (a + 10d) \]
This simplifies to:
\[ Q = 6a + (0 + 2 + 4 + 6 + 8 + 10)d = 6a + 30d \]

### Step 5: Use the ratio \( P:Q = 32:27 \)
Given that \( P:Q = 32:27 \), we can write:
\[ \frac{P}{Q} = \frac{32}{27} \]
Substituting for \( P \) and \( Q \):
\[ \frac{6a + 36d}{6a + 30d} = \frac{32}{27} \]
Cross multiplying gives:
\[ 27(6a + 36d) = 32(6a + 30d) \]
Expanding both sides:
\[ 162a + 972d = 192a + 960d \]
Rearranging gives:
\[ 162a - 192a + 972d - 960d = 0 \]
This simplifies to:
\[ -30a + 12d = 0 \quad \Rightarrow \quad 30a = 12d \quad \Rightarrow \quad 5a = 2d \quad \Rightarrow \quad a = \frac{2d}{5} \quad \text{(Equation 2)} \]

### Step 6: Substitute Equation 2 into Equation 1
Now substitute \( a = \frac{2d}{5} \) into Equation 1:
\[ 2\left(\frac{2d}{5}\right) + 11d = 59 \]
This simplifies to:
\[ \frac{4d}{5} + 11d = 59 \]
Multiplying through by 5 to eliminate the fraction:
\[ 4d + 55d = 295 \]
Combining terms gives:
\[ 59d = 295 \]
Solving for \( d \):
\[ d = \frac{295}{59} = 5 \]

### Conclusion
The common difference \( d \) of the arithmetic progression is:
\[ \boxed{5} \]