Question

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 1

Let the original numbers be 

a, a + d, and a + 2d 

According to the question

a + a + d + a + 2d = 15 

⇒ 3a + 3d = 15 or a + d = 5 

⇒ d = 5 – a 

After the addition, the three numbers are: 

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

they are now in GP, that is –

\(\Rightarrow\)\(\frac{a + d + 3}{a + 1}\) = \(\frac{a + 2d + 9 }{a + d+ 3}\)

⇒ (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 

∴ The numbers are 3,5,7 or 15,5, – 5