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