Question

Find the smallest of the three numbers in arithmetic progression, if the product
of the first and the third numbers is 252 and the sum of the three numbers is 48.
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Answer 1

Answer:

Step-by-step explanation:

Let the 3 no in A.P be : a−d,a,a+d

∴ Sum of these number =(a−d)+(a)+a+d

12=3a

4=a

and also

a+d=3(a−d)

a+d=3a−3d

4d=2a

2d=a

d=4/2=2

d=2

∴A.P⇒a−d,a,a+d

⇒4−2,4,4+2

⇒2,4,6.

∴ product ⇒2×4×6

⇒48 (option D)