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) ⇒ Sn = 1 + 5 + 12 + 22 + 35 + … + an + 0 ------(3)
⇒ Sn = 0 + 1 + 5 +12 + 22 + 35 + … an-1 + an -----(4)
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(3) - (4) ⇒ 0 = 1 + 4 + 7 + 10 + 13 + … + (an - an-1) - an
⇒ an = 1 + 4 + 7 + 10 + 13 + … upto n terms
⇒ an = 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., Sn = n/2 (2a + (n - 1)d), a = 4, d = 3 and n = n - 1
∴ an = 1 + {(n - 1) / 2 [(2 × 4) + (n - 1 - 1) 3]}
an = 1 + {(n - 1) / 2 [8 + 3n - 6]}
an = {2 + [(n - 1) (3n + 2)]} / 2
∴ Required nth term is an = {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
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.
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