Note: The sum of the series is already provided in the question. The solution to find x is given below.

Let there be n terms in the series.

x = 1 + (n - 1)3

= 3n - 2

Let S be the sum of the series

⇒n[1 + 3n - 2] = 1430

⇒n + 3n² - 2n = 1430

⇒3n² - n - 1430 = 0

Applying Sri Dhar Acharya formula, we get

⇒ n = 22 as n cannot be a fraction

Therefore x = 3 × 22 - 2 = 64

The value of x is 64