Question

If the n^th terms of the two APs: 9, 7, 5,... and 24, 21, 18,... are the same, find the value of n. Also find that term.

Answer 1

Let the first term, common difference and number of terms of the AP: 9, 7, 5,... are a₁, d₁ and n₁, respectively.

i.e., first term (a₁) = 9

And common difference (d₁) = 7 – 9 = – 2

∴ Its n^th term,

⇒ `T_(n_1)^(')` = a₁ + (n₁ – 1)d₁

⇒ `T_(n_1)^(')` = 9 + (n₁ – 1)(– 2)

⇒ `T_(n_1)^(')` = 9 – 2n₁ + 2

⇒ `T_(n_1)^(')` = 11 – 2n₁  [∵ n^th term of an AP, T_n = a + (n – 1)d]...(i)

Let the first term, common difference and the number of terms of the AP: 24, 21, 18,... are a₂, d₂ and n₂, respectively.

i.e., first term, (a₂) = 24

And common difference (d₂) = 21 – 24 = – 3

∴ Its n^th term,

`T_(n_2)^('')` = a₂ + (n₂ – 1)d₂

⇒ `T_(n_2)^('')` = 24 + (n₂ – 1)(– 3)

⇒ `T_(n_2)^('')` = 24 – 3n₂ + 3

⇒ `T_(n_2)^('')` = 27 – 3n₂  ...(ii)

Now, by given condition,

n^th terms of the both APs are same, 

i.e., `T_(n_1)^(') = T_(n_2)^('')`

11 – 2n₁ = 27 – 3n₂   ...[From equations (i) and (ii)]

⇒ n = 16 

∴ n^th term of first AP,

`T_(n_1)^(')` = 11 – 2n₁

= 11 – 2(16)

= 11 – 32

= – 21

And n^th term of second AP,

`T_(n_2)^('')` = 27 – 3n₂

= 27 – 3(16)

= 27 – 48

= – 21

Hence, the value of n is 16 and that term i.e., n^th term is – 21.