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.

#### Solution

Let the first term, common difference

And number of terms of the AP: 9, 7, 5,…. are

a_{1}, d_{1} and n_{1}, respectively

i.e., first term (a_{1}) = 9

And common difference (d_{1}) = 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 a_{2}, d_{2} and n_{2}, respectively

i.e., first term, (a_{2}) = 24

And common difference (d_{2}) = 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.