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

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