Q. 4 A5.0( 6 Votes )

Sum of three numbers in A.P. is 21 and their product is 231. Find the numbers.

Answer :

Let the three numbers are (a – d), a and (a + d)


According to question,


Sum of these three numbers = 21


a – d + a + a + d = 21


3a = 21


a = 7 …(i)


and it is also given that


Product of these numbers = 231


(a – d) × a × (a + d) = 231


(7 – d) × 7 × (7 + d) = 231


7 × (72 – d2) = 231 [ (a – b)(a + b) = a2 – b2]


7 × (49 – d2) = 231


343 – 7d2 = 231


– 7d2 = 231 – 343


– 7d2 = – 112


d2 = 16


d = √16


d = ±4


Case I: If d = 4 and a = 7


a – d = 7 – 4 = 3


a = 7


a + d = 7 + 4 = 11


So, the numbers are


3, 7, 11


Case II: If d = – 4 and a = 7


a – d = 7 – ( – 4) = 7 + 4 = 11


a = 7


a + d = 7 + ( – 4) = 7 – 4 = 3


So, the numbers are


11, 7, 3


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