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

⇒ `T"'"_(n_1) = a_1 + (n_1 - 1)d_1`

⇒ `T"'"_(n_1) =9 + (n_1 - 1)(-2)`

⇒ `T"'"_(n_1) = 9 - 2n_1 + 2`

⇒ `T"'"_(n_1) = 11 - 2n_1`   ......(i)

∵ n^th term of an AP, Tn = a + (n – 1)d

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) = a_2 + (n_2 - 1)d_2`

⇒ `T"'"_(n_2) = 24 + (n_2 - 1)(-3)`

⇒ `T"'"_(n_2) = 24 - 3n_2 + 3`

⇒ `T"'"_(n_2) = 27 - 3n_2`  .....(ii)

Now, by given condition,

n^th terms of the both APs are same 

i.e., `11 - 2n_1 = 27 - 3n_2`   ......[From equations (i) and (ii)]

⇒ n = 16  ......`[because n_1 - n_2 = n]`

∴ n^th terms of first AP,

`T"'"n_1 = 11 - 2n_1`

= `11 - 2(16)`

= `11 - 32`

= – 21

And n^th terms of second AP,

`T"'"n_2 = 27 - 3n_2`

= `27 - 3(16)`

= `24 - 48`

= – 21

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