Find four consecutive terms in an A.P. Whose sum is 72 and the ratio of the product of the 1st and the 4th terms to the product of the 2nd and 3rd terms is 9 : 10.
(a) Let the four consecutive terms be a - 3d, a - d, a + d, a + 3d.
(b) Using the first condition, find the value of a.
(c) Using the second condition, find the value of d and hence the numbers.

Answer

To solve the problem, we will follow the steps outlined in the video transcript to find the four consecutive terms in an arithmetic progression (A.P.) whose sum is 72 and the ratio of the product of the 1st and 4th terms to the product of the 2nd and 3rd terms is 9:10.

### Step 1: Define the terms
Let the four consecutive terms in the A.P. be:
- First term: \( a - 3d \)
- Second term: \( a - d \)
- Third term: \( a + d \)
- Fourth term: \( a + 3d \)

### Step 2: Use the first condition (sum of the terms)
According to the problem, the sum of these four terms is 72:
\[ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 72 \]
Simplifying this, we get:
\[ 4a = 72 \]
Dividing both sides by 4:
\[ a = 18 \]

### Step 3: Use the second condition (ratio of products)
The second condition states that the ratio of the product of the 1st and 4th terms to the product of the 2nd and 3rd terms is 9:10. We can express this mathematically:
\[ \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{9}{10} \]
Calculating the products:
- First and fourth terms:
\[ (a - 3d)(a + 3d) = a^2 - 9d^2 \]
- Second and third terms:
\[ (a - d)(a + d) = a^2 - d^2 \]
Substituting these into the ratio gives:
\[ \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{9}{10} \]

### Step 4: Substitute \( a = 18 \)
Now substituting \( a = 18 \):
\[ \frac{18^2 - 9d^2}{18^2 - d^2} = \frac{9}{10} \]
Calculating \( 18^2 \):
\[ \frac{324 - 9d^2}{324 - d^2} = \frac{9}{10} \]

### Step 5: Cross-multiply to solve for \( d \)
Cross-multiplying gives:
\[ 10(324 - 9d^2) = 9(324 - d^2) \]
Expanding both sides:
\[ 3240 - 90d^2 = 2916 - 9d^2 \]
Rearranging the equation:
\[ 3240 - 2916 = 90d^2 - 9d^2 \]
\[ 324 = 81d^2 \]
Dividing both sides by 81:
\[ d^2 = 4 \]
Taking the square root:
\[ d = 2 \]

### Step 6: Find the four terms
Now we can find the four consecutive terms using \( a = 18 \) and \( d = 2 \):
1. First term: \( a - 3d = 18 - 6 = 12 \)
2. Second term: \( a - d = 18 - 2 = 16 \)
3. Third term: \( a + d = 18 + 2 = 20 \)
4. Fourth term: \( a + 3d = 18 + 6 = 24 \)

### Final Answer
The four consecutive terms in the A.P. are:
\[ 12, 16, 20, 24 \]