Q. 4 A5.0( 6 Votes )

# 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

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