for '$n$', last position should be $\frac{n(n+1)}{2}$ and first position should be $\frac{n(n+1)}{2} -n+1$. right ?

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+2 votes

Given the sequence $A,B,B,C,C,C,D,D,D,D,\ldots$ etc$.,$ that is one $A,$ two $B’s,$ three $C’s,$ four $D’s,$ five $E’s$ and so on, the $240^{th}$ latter in the sequence will be $:$

- $V$
- $U$
- $T$
- $W$

+1 vote

Best answer

Given Sequence is :

$A,B,B,C,C,C,D,D,D,D,\ldots$ etc.

Here, each alphabet is repeated as many times as its position in the alphabetical order. $$\begin{array}{|c|c|l|} \hline \textbf{Alphabet} & \textbf{Position in} & \textbf{Position in the Given Sequence} \\& \textbf{alphabetical order}&\\\hline A & 1 & 1^{st} \\\hline B & 2 & 2^{nd} \text{ and }\ \ 3^{rd} \\\hline C & 3 & 4^{th}, 5^{th} \text{ and } \ \ 6^{th} \\\hline D & 4 & 7^{th}, 8^{th}, 9^{th} \text{ and } \ \ 10^{th} \\\hline \ldots & \ldots& \ldots \\\hline \text{Some Alphabet} & n & \left ( \frac{n(n+1)}{2} - n+1 \right ),

\left ( \frac{n(n+1)}{2} - n+2 \right ),\\&&\ldots,240^{th},\ldots, \left(\frac{n.(n+1)}{2}\right)\\

\hline

\end{array}$$ Here, $\frac{n(n+1)}{2}$ is the last position in the given sequence for an alphabet whose position is $n$ in the alphabetical order. So, If we have to find an alphabet whose position is $i^{th}$ in the given sequence and whose position is $n$ in the alphabetical order then we have to find smallest integer value of $n$ such that $\frac{n(n+1)}{2} \geqslant i$

For example, if $i=5,$ it means we have to find alphabet whose position is $5^{th}$ in the given sequence, then we have to find minimum value of $n$ such that $\frac{n(n+1)}{2} \geqslant 5$, So, minimum value of $n = 3$ means it will be alphabet $C.$

Similarly, if $i=8$ means we have to find alphabet whose position is $8^{th}$ in the given sequence, then we have to find minimum value of $n$ such that $\frac{n(n+1)}{2} \geqslant 8$, So, minimum value of $n = 4$ means it will be alphabet $D.$

Now, in this question, we have to find **smallest** integer value of $n$ such that

$\quad\frac{n(n+1)}{2}$ $\geq$ $240$ $(\because$ Sum of first $n$ natural numbers = $\frac{n(n+1)}{2})$

$\quad \Rightarrow n*(n+1)\geq 480$

Options given are $V, U, T, W.$ So, value of $n$ can be $22,21,20$ or $23.$

$22*23 = 506$ $\geq$ $480$ $\Rightarrow$ $n= 22.$

The $22^{nd}$ alphabet is $V.$

$\therefore$ Option $(A).$ $V$ is the correct answer.

+1 vote

$A_1|B_2|B_3|C_4|C_5|C_6|D_7|D_8|D_9|D_{10}|E|E|E|E|E|......Z|Z|Z$

$A :sum=1$

$BB:sum=1+2=3$

$CCC:sum=1+2+3=6$

$DDDD:sum=1+2+3+4=10$

**The letter D occupies the $4^{th}$ position and it gives a sum of $10$ by $\frac{n(n+1)}{2},$ means the last D will be the $10^{th}$ term and the first D will be the (10-4+1)=$7^{th}$ term.**

The letter $V$ occupies the $22^{nd}$ position

$\therefore$ the last $V$ will be the $\frac{22\times 23}{2} = 253^{rd}$ term

The first $V$ will be the $253 - 22 + 1 = 232^{nd}$ term

$V$ lies b/w $232$ to $253$

**Similarly you can check for other options to.**

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