Question

Two A.P.’s are given 9, 7, 5, ... and 24, 21, 18, ... If n^th term of both the progressions are equal then find the value of n and n^th term.

Answer 1

The given sequence is 9, 7, 5, ...

Here, a = 9, d = 7 – 9 = –2

t_n = a + (n – 1)d    ...(Formula)

= 9 + (n – 1) × (–2)

= 9 – 2n + 2

∴ t_n = 11 – 2n    ...(1)

The second A.P. is 24, 21, 18, ...

Here, a = 24, d = 21 – 24 = –3

t_n = a + (n – 1)d    ...(Formula)

= 24 + (n – 1) × (–3)

= 24 – 3n + 3

∴ t_n = 27 – 3n    ...(2)

From the given condition,

11 – 2n = 27 – 3n

∴ –2n + 3n = 27 – 3n

∴ n = 16

Now we find t₁₆

t_n = a + (n – 1)d    ...(Formula)

∴ t₁₆ = 9 + (16 – 1) × (–2)    ...(Taking values from 1^st A.P.)

= 9 + 15 × (–2)

= 9 – 30

∴ t₁₆ = –21

∴ The value of n is 16, and the n^th (16^th) term is –21.