Let an = 89
AP = 4, 9, 14, …89
Here, a = 4, d = 14 – 9 = 5
We know that
an = a + (n – 1)d
⇒ 89 = 4 + (n – 1)5
⇒ 85 = (n – 1)5
⇒ 17 = n – 1
⇒ 18 = n
So, 89 is the 18th term of the given AP
Now, we find the sum of 4 + 9 + 14 + … + 89
We know that,
⇒ S18 = 9[8 + 17 × 5]
⇒ S18 = 9[8 + 85]
⇒ S18 = 9 × 93
⇒ S18 = 837
Hence, the sum of the given AP is 837.