Question

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

Answer 1

$122^n^d$ term will be 16.

Step-by-step explanation:

We are given a series as;

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

In the above series, we observe that;

• $1^s^t$ term = 1

• $2^n^d$ and $3^r^d$ term = 2

• $4^t^h$, $5^t^h$ and$6^t^h$ term = 3

• $7^t^h$, $8^t^h$, $9^t^h$ and $10^t^h$ term = 4

• $11^t^h$, $12^t^h$, $13^t^h$, $14^t^h$ and $15^t^h$ = 5

• $16^t^h$, $17^t^h$, $18^t^h$, $19^t^h$ and $20^t^h$ = 6

• Therefore, if make a sequence by ending terms of each group, we get as;

1, 3, 6, 10, 15, 20, ....................

In general this series can be represented by;

• $\dfrac{n(n + 1)}{2}$      where; n = 1, 2, 3, 4, 5, .........

Using this general form, we can find $122^n^d$ term, very easily.

• For eg: we put n = 5 in $\dfrac{5(5 + 1)}{2}$

       = 15

• It means that; $15^t^h$ term is 5.

• Now, find out the greatest integer value of 'n', for which $\dfrac{n(n + 1)}{2} < 122$

Therefore, n = 15

• $\dfrac{n(n + 1)}{2} < 122$

120 < 122

• Thus, $120^n^d$ term will be 15

It means that$122^n^d$ term will be 16.