What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$ 
What is the 94th term of the following sequence?
  $$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
  
  
*
  
*8
  
*9
  
*10
  
*11

My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is written 2n: n is natural number. Here  94 = 2(47)  so sum of first few natural numbers should be greater than or equal to 47. Since $$1+2+3+4+5+6+7+8+9 = 45 < 47$$  so options 1,2 are not possible and $$1+2+3+4+5+6+7+8+9+10 = 55 >47$$ But this is a lengthy process.
Please tell me easiest way to approach the answer.
 A: Look at the last numbers of the repeating numbers. Notice the arithmetic sequence: $2,4,6,...,2+2(n-1)$, whose sum of $n$ terms is: $S_n=(n+1)n$. So, the general formula is: $a_{S_n}=n$. For example:
$$a_{S_\color{red}1}=a_2=\color{red}1\\
a_{S_\color{red}2}=a_6=\color{red}2\\
\vdots\\
a_{S_n}=a_{94}=?$$
We make up the equation:
$$(n+1)n=94 \Rightarrow n^2+n-94=0 \Rightarrow n\approx 9.2>9 \Rightarrow n=10.$$
Verify:
$$S_9=(9+1)\cdot 9=90 \Rightarrow a_{90}=9\\
S_{10}=(10+1)\cdot 9=99 \Rightarrow a_{99}=10$$
So, $a_{91}=a_{92}=\cdots=a_{99}=10$.
A: Using your method,
$n(n+1) \lt 94$ 
Easy to see $9*10 = 90$ so the $90$th term value is $9$.
And $91$st term is the beginning of value $10$.  
$a(a+1)$th term has the value $a$ and its the end of that value.
A: To detremine the $94$-th term, I can use the rules of the aritmethic progression, in particular: $$S_n=\frac{n}{2}(2a_0+(n-1)d)$$ Substituing the numbers, I have: $$S_n=\frac{n}{2}(2+2n)$$ Imposig $S_n=94$, I obtain: $n^2+n-94=0$ and so $n=10$.
