Question

122nd term of the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ........ Is

Answer 1

it is 122 i get it by using Tn =a +(n-1)d
in it a =1,d=1,n=122

Answer 2

122nd term of sequence is 16

Solution:

Given sequence is:

1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 4 , .........

To find: 122nd term of sequence

The formula for the nth term of the sequence where, "n" occurs n times is:

$n\ th\ term = \frac{ 1 + \sqrt{1+8n}}{2}$

We have to find 122nd term

n = 122

Then,

$122nd\ term = \frac{1 + \sqrt{1 + 8(122)}}{2} \\\\122nd\ term = \frac{1 +\sqrt{977}}{2}\\\\122nd\ term = 16.12 \approx 16$

Thus 122nd term of sequence is 16

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