Question

The sum of 3 numbers in A.P. is 48. The sum of their squares is 800. Find the smallest number among the three numbers.

• A. \(10 \)
• B. \(12 \)
• C. \(11 \)
• D. \(9 \)
• E. \(13 \)

Hint:

For 3 numbers in A.P.: - Represent as $a-d, a, a+d$ - It simplifies both sum and square calculations

Answer 1

The Correct Option is A

Concept: Three numbers in A.P. can be written as: \[ a-d, \quad a, \quad a+d \]
• Sum = $3a$
• Sum of squares = $(a-d)^2 + a^2 + (a+d)^2$
Step 1: Use sum condition.
\[ (a-d) + a + (a+d) = 48 \] \[ 3a = 48 \Rightarrow a = 16 \]

Step 2: Use sum of squares.
\[ (a-d)^2 + a^2 + (a+d)^2 = 800 \] Substitute $a = 16$: \[ (16-d)^2 + 16^2 + (16+d)^2 = 800 \] Expand: \[ (256 - 32d + d^2) + 256 + (256 + 32d + d^2) = 800 \] \[ 768 + 2d^2 = 800 \]

Step 3: Solve for $d$.
\[ 2d^2 = 32 \Rightarrow d^2 = 16 \Rightarrow d = 4 \]

Step 4: Find smallest number.
\[ \text{Smallest} = a - d = 16 - 4 = 12 \]