Question

there numbers are in an arithmetical progression and there sum is 15. If 1,3,9 respectively, be added to them, they from a geometrical progression. Find the numbers.

Answer 1

Answer :

Let us consider that the numbers in AP are
(a - d), a and (a + d).

By the given condition,

(a - d) + a + (a + d) = 15

⇒ 3a = 15

⇒ a = 5

So, the numbers are determined as
(5 - d), 5 and (5 + d).

Now, 1, 3 and 9 are added to (5 - d), 5 and (5 + d); so the numbers are

(5 - d) + 1, 5 + 3, (5 + d) + 9

i.e., (6 - d), 8, (14 + d).

Given that, this numbers be in GP.

Then,

8² = (6 - d)(14 + d)

⇒ 64 = 84 + 6d - 14d - d²

⇒ 64 = 84 - 8d - d²

⇒ d² + 8d + 64 - 84 = 0

⇒ d² + 8d - 20 = 0

⇒ d² + 10d - 2d - 20 = 0

⇒ d (d + 10) - 2 (d + 10) = 0

⇒ (d + 10)(d - 2) = 0

∴ d + 10 = 0 & d - 2 = 0

⇒ d = - 10 & d = 2

∴ the numbers are

5 - (- 10), 5, 5 + (- 10); 5 - 2, 5, 5 + 2

i.e., (15, 5, - 5); (3, 5, 7)

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