Question

The ${2008}^{th}$ term of the sequence $1,\underset{3}{\underset{}{2,2,2,}}\underset{6}{\underset{}{3,3,3,3,3,3,}}\underset{10}{\underset{}{4,4,4,4,4,4,4,4,4,4,}}...$ where n occurs $\frac{n(n+1)}{2}$ times in the sequence, equals.

• A. 24
• B. 23
• C. 22
• D. 21

Answer 1

The correct option is C 22

No. of terms
Group (1) 1 1
Group (2) 2, 2, 2 3
Group (3) 3, 3, ..., 3 6
Group (4) 4, 4, ..., 4 10
. . . . .
. . . . .
Group (r) r, r, ..., r $\frac{r^2+r}{2}$
Let ${2008}^{th}$ term falls in $r^{th}$ group
$\Rightarrow 1+3+6+10+...+\frac{(r-1{)}^2+(r-1)}{2}<2008\le 1+3+6+...+\frac{r^2+r}{2}$
$\Rightarrow \frac{(r-1)r(r+1)}{6}<2008\le \frac{r(r+1)(r+2)}{6}$
$\Rightarrow r^3-r<12048\le (r+1{)}^3-(r+1)....(i)$
$\Rightarrow$ r will be nearer to cube root of 12048
Note:$22<\sqrt[3]{312048}<23$
for r=22 inequality (i) holds
for r< 22 RHS of (1) is less than 12048
for r $\ge$ 23 LHS of (1) is greater than 12048
$\Rightarrow r=22$ is the required value $\Rightarrow {2008}^{th}$ term is 22