Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to

1.	Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.
Answer» Let the first term of an A.P. be ‘a’ and its common difference be‘d’. a₁ + a₂ + a₃ = 15 Where, the three number are: a, a + d, and a + 2d So, a + a + d + a + 2d = 15 3a + 3d = 15 or a + d = 5 d = 5 – a … (i) Now, according to the question: a + 1, a + d + 3, and a + 2d + 9 they are in GP, that is: (a + d + 3)/(a + 1) = (a + 2d + 9)/(a + d + 3) (a + d + 3)² = (a + 2d + 9) (a + 1) a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9 (5 – a)² – 4a + 4(5 – a) = 0 25 + a² – 10a – 4a + 20 – 4a = 0 a² – 18a + 45 = 0 a² – 15a – 3a + 45 = 0 a(a – 15) – 3(a – 15) = 0 a = 3 or a = 15 d = 5 – a d = 5 – 3 or d = 5 – 15 d = 2 or – 10 Then, For a = 3 and d = 2, the A.P is 3, 5, 7 For a = 15 and d = -10, the A.P is 15, 5, -5 ∴ The numbers are 3, 5, 7 or 15, 5, – 5

Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.

Answer»

Let the first term of an A.P. be ‘a’ and its common difference be‘d’.

a₁ + a₂ + a₃ = 15

Where, the three number are: a, a + d, and a + 2d

So,

a + a + d + a + 2d = 15

3a + 3d = 15 or a + d = 5

d = 5 – a … (i)

Now, according to the question:

a + 1, a + d + 3, and a + 2d + 9

they are in GP, that is:

(a + d + 3)/(a + 1) = (a + 2d + 9)/(a + d + 3)
(a + d + 3)² = (a + 2d + 9) (a + 1)

a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9

(5 – a)² – 4a + 4(5 – a) = 0

25 + a² – 10a – 4a + 20 – 4a = 0

a² – 18a + 45 = 0

a² – 15a – 3a + 45 = 0

a(a – 15) – 3(a – 15) = 0

a = 3 or a = 15

d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

Then,

For a = 3 and d = 2, the A.P is 3, 5, 7

For a = 15 and d = -10, the A.P is 15, 5, -5

∴ The numbers are 3, 5, 7 or 15, 5, – 5