Question

The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Answer 1

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

a₁ + a₂ + a₃ = 21

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

So,

3a + 3d = 21 or

a + d = 7.

d = 7 – a …. (i)

Now, according to the question:

a, a + d – 1, and a + 2d + 1

they are now in GP, that is:

(a + d - 1)/a = (a + 2d + 1)/(a + d - 1)

(a + d – 1)² = a(a + 2d + 1)

a² + d² + 1 + 2ad – 2d – 2a = a² + a + 2da

(7 – a)² – 3a + 1 – 2(7 – a) = 0

49 + a² – 14a – 3a + 1 – 14 + 2a = 0

a² – 15a + 36 = 0

a² – 12a – 3a + 36 = 0

a(a – 12) – 3(a – 12) = 0

a = 3 or a = 12

d = 7 – a

d = 7 – 3 or d = 7 – 12

d = 4 or – 5

Then,

For a = 3 and d = 4, the A.P is 3, 7, 11

For a = 12 and d = -5, the A.P is 12, 7, 2

∴ The numbers are 3, 7, 11 or 12, 7, 2