Question

Three numbers are in arithmetic progression (A.P.). Their sum is 21 and the product of the first number and the third number is 45. Then the product of these three numbers is:

• A. 315
• B. 90
• C. 180
• D. 270
• E. 450

Hint:

Once you find the middle term (\( a \)) and the product of the extremes (\( a^2 - d^2 \)), the total product is simply \( a \times (a^2 - d^2) \). Here, \( 7 \times 45 = 315 \). This shortcut saves you from finding \( d \) and the individual numbers.

Answer 1

The Correct Option is A

Concept: To find the product of three numbers in A.P., we first determine the individual terms. As established, assuming the terms are \( a-d \), \( a \), and \( a+d \) is the most efficient algebraic approach for odd-numbered sets in an A.P.

Step 1: Determining the middle term and the common difference.
The sum of the three numbers is 21: \[ (a-d) + a + (a+d) = 21 \implies 3a = 21 \implies a = 7 \] We are given that the product of the first and third numbers is 45: \[ (a-d)(a+d) = 45 \implies a^2 - d^2 = 45 \] Substitute \( a = 7 \): \[ 7^2 - d^2 = 45 \implies 49 - d^2 = 45 \] \[ d^2 = 4 \implies d = 2 \text{ (or } -2 \text{)} \]

Step 2: Calculating the product of the three numbers.
Using \( a = 7 \) and \( d = 2 \), the numbers are \( (7-2), 7, (7+2) \), which are \( 5, 7, 9 \). The product of these three numbers is: \[ \text{Product} = 5 \times 7 \times 9 \] \[ \text{Product} = 35 \times 9 = 315 \]