What is the 50th term of the sequence whose nth term is given by a formula in terms of n, and the first five terms are 1, 5, 12, 22, 35?

Solution:

Given sequence is 1, 5, 12, 22, 35,... ------(1)

Difference between the terms is 4, 7, 10, 13,... -------(2)

We find that differences between the terms form an arithmetic progression.

∴ consider the series

For (1) ⇒ S_n = 1 + 5 + 12 + 22 + 35 + … + a_n + 0       ------(3)

            ⇒ S_n = 0 + 1 + 5 +12 + 22 + 35 + … a_n-1 + a_n -----(4)

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(3) - (4) ⇒ 0 = 1 + 4 + 7 + 10 + 13 + … + (a_n - a_n-1) - a_n

⇒ a_n = 1 + 4 + 7 + 10 + 13 + … upto n terms

⇒ a_n = 1 + [4 + 7 + 10 + 13 + … upto (n - 1)] terms

 where [4 + 7 + 10 + 13 + … upto (n - 1)] 

⇒ sum of (n -1) terms of A.P., S_n = n/2 (2a + (n - 1)d), a = 4, d = 3 and n = n - 1

∴ a_n = 1 + {(n - 1) / 2 [(2 × 4) + (n - 1 - 1) 3]}

a_n = 1 + {(n - 1) / 2 [8 + 3n - 6]}

a_n = {2 + [(n - 1) (3n + 2)]} /  2

∴ Required nth term is a_n = {2 + [(n - 1) (3n + 2)]} / 2

To find 50th term, substitute n = 50 in the above equation

a₅₀ = {2 + [(50 - 1) ((3 × 50) + 2)]} / 2 

a₅₀ = 3725

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What is the 50th term of the sequence whose nth term is given by a formula in terms of n, and the first five terms are 1, 5, 12, 22, 35?

Summary:

The 50th term of the sequence whose nth term is given by a formula in terms of n, and the first five terms are 1, 5, 12, 22, 35, is 3725.