3

What is the 94th term of the following sequence? $$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$

  1. 8

  2. 9

  3. 10

  4. 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.

Blue
  • 75,673
Jyoti bagotia
  • 49
  • I see that as being the easiest way of solving it. Did you use the formula $1+2+3+4+...+n = \frac{n(n+1)}{2}$? – Rick Aug 31 '19 at 03:52
  • Yes, I used the above formula. – Jyoti bagotia Aug 31 '19 at 03:59
  • 2
    If that's the formula and technique you used, I don't really see much more that can be said. It's how I'd do it, and it only takes a few minutes at most if you're quick to pick up on what you should do. Perhaps there is some more elaborate way to do it, e.g. finding a formula for the sequence, but if you get an idea on how to work it it might be best to pursue that idea, as opposed to looking for the "most efficient" idea and wasting time in just trying to figure out how to waste less time. – PrincessEev Aug 31 '19 at 04:02
  • 1
    FWIW though, the OEIS has an entry for this sequence (with a zero appended to the start): https://oeis.org/A000194. Several formulas are given. One of the most attractive is the one below, but I have no clue how to derive it:

    $$a(n) = \left \lfloor \sqrt n + \frac 1 2 \right \rfloor$$

    where $n$ is the position of the sequence whose term you want, and $\lfloor \cdot \rfloor$ denotes the floor ("least integer") function.

     – PrincessEev Aug 31 '19 at 04:09

3 Answers3

3

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, $a_{91}=a_{92}=\cdots=a_{99}=10$.

farruhota
  • 31,482
  • Judging from the OEIS entry cited in the comments above, this procedure can evidently be sharpened to: $n\approx \sqrt{94}=9.7\implies n=10$. – Semiclassical Aug 31 '19 at 04:32
2

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.

AgentS
  • 12,195
1

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 $n=10$.

Matteo
  • 6,581