Three numbers are in AP, and their sum is 15. If 1, 4, 19 be added to

1.	Three numbers are in AP, and their sum is 15. If 1, 4, 19 be added to them respectively, then they are in GP. Find the numbers.
Answer» To find: The numbers Given: Three numbers are in A.P. Their sum is 15 Formula used: When a,b,c are in GP, b² = ac Let the numbers be a - d, a, a + d According to first condition a + d + a +a – d = 15 ⇒ 3a = 15 ⇒ a = 5 Hence numbers are 5 - d, 5, 5 + d When 1, 4, 19 be added to them respectively then the numbers become – 5 – d + 1, 5 + 4, 5 + d + 19 ⇒ 6 – d, 9, 24 + d The above numbers are in GP Therefore, 9² = (6 – d) (24 + d) ⇒ 81 = 144 – 24d +6d – d 2 ⇒ 81 = 144 – 18d – d 2 ⇒ d 2 + 18d – 63 = 0 ⇒ d 2 + 21d – 3d – 63 = 0 ⇒ d (d + 21) -3 (d + 21) = 0 ⇒ (d – 3) (d + 21) = 0 ⇒ d = 3, Or d = -21 Taking d = 3, the numbers are 5 - d, 5, 5 + d = 5 - 3, 5, 5 + 3 = 2, 5, 8 Taking d = -21, the numbers are 5 - d, 5, 5 + d = 5 – (-21), 5, 5 + (-21) = 26, 5, -16 We have two sets of triplet as 2, 5, 8 and 26, 5, -16

Three numbers are in AP, and their sum is 15. If 1, 4, 19 be added to them respectively, then they are in GP. Find the numbers.

Answer»

To find:

The numbers Given:

Three numbers are in A.P. Their sum is 15

Formula used:

When a,b,c are in GP, b² = ac

Let the numbers be a - d, a, a + d

According to first condition

a + d + a +a – d = 15

⇒ 3a = 15

⇒ a = 5

Hence numbers are 5 - d, 5, 5 + d

When 1, 4, 19 be added to them respectively then the numbers become –

5 – d + 1, 5 + 4, 5 + d + 19

⇒ 6 – d, 9, 24 + d

The above numbers are in GP

Therefore, 9² = (6 – d) (24 + d)

⇒ 81 = 144 – 24d +6d – d 2

⇒ 81 = 144 – 18d – d 2

⇒ d 2 + 18d – 63 = 0

⇒ d 2 + 21d – 3d – 63 = 0

⇒ d (d + 21) -3 (d + 21) = 0

⇒ (d – 3) (d + 21) = 0

⇒ d = 3, Or d = -21

Taking d = 3, the numbers are

5 - d, 5, 5 + d = 5 - 3, 5, 5 + 3

= 2, 5, 8

Taking d = -21, the numbers are 5 - d, 5, 5 + d = 5 – (-21), 5, 5 + (-21)

= 26, 5, -16

We have two sets of triplet as 2, 5, 8 and 26, 5, -16