Answer :

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


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