Which term of the A.P.

(i) 3, 8, 13, 18, … is 78 ?

(ii) $18, 15\dfrac{1}{2}, 13, …$ is -47 ?

(i) The given A.P. is 3, 8, 13, 18 ….

Here, first term = a = 3 and common difference = d = 8 - 3 = 5.

Let 78 be nth term of the A.P., or a_n = 78

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of d and a_n in above equation we get,

⇒ a_n = a + (n - 1)d
⇒ 78 = 3 + (n - 1)5
⇒ 78 = 3 + 5n - 5
⇒ 78 = 5n - 2
⇒ 5n = 78 + 2
⇒ 5n = 80
⇒ n = 16.

Hence, 78 is 16th term of the A.P.

(ii) The given A.P. is $18, 15\dfrac{1}{2}, 13, …$

Here, first term = a = 18 and common difference = d = $\dfrac{31}{2} - 18 = \dfrac{31 - 36}{2} = -\dfrac{5}{2}$.

Let -47 be nth term of the A.P., or a_n = -47

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of a, d and a_n in above equation we get,

$a_n = a + (n - 1)d \\[1em] \Rightarrow -47 = 18 + (n - 1)\Big(-\dfrac{5}{2}\Big) \\[1em] \Rightarrow -47 = 18 -\dfrac{5n}{2} + \dfrac{5}{2} \\[1em] \Rightarrow -47 - 18 = \dfrac{5 - 5n}{2} \\[1em] \Rightarrow -65 \times 2 = 5 - 5n \\[1em] \Rightarrow -130 = 5 - 5n \\[1em] \Rightarrow 5n = 130 + 5 \\[1em] \Rightarrow 5n = 135 \\[1em] \Rightarrow n = 27.$

Hence, -47 is 27th term of the A.P.