Find n so that the nth terms of the following two A.P.’s are the same.

1, 7, 13, 19,… and 100, 95, 90,… .

For first A.P. , first term, a = 1; common difference, d = 7 – 1 = 6

For second A.P, first term, a = 100; common difference , d = 95 – 100 = -5

We know that nth term of A.P. , t_n = a + (n – 1) d.

First A.P:

⇒ t_n1 = 1 + (n – 1) (6)

= 1 + 6n – 6

∴ t_n1 = 6n – 5

Second A.P:

⇒ t_n2 = 100 + (n – 1) (-5)

= 100 – 5n + 5

∴ t_n2 = 105 – 5n

Given that nth terms of the two A.P.’s are the same.

∴ t_n1 = t_n2

⇒ 6n – 5 = 105 – 5n

⇒ 6n + 5n = 105 + 5

⇒ 11n = 110

⇒ n = 110/11 = 10

∴ For the nth terms of the two given A.P.’s to be same, the value of n = 10.