Pentagonal Numbers - Reversing The Pentagon

By the generalized formula, we have

$1001 = \frac{3n^2-n}{2}$
$\Rightarrow 3n^2-n-2002=0$

Solving this quadratic equation, we get

$n = \frac{-77}{3}$ or $n = 26$ .

Since $n$ is a natural number,

$n = \boxed{26}$ .

We need to use the formula of $\frac{3n^{2}-n}{2}$ to find the $n^{th}$ pentagonal figure.

$\frac{3n^{2}-n}{2}=1001$

$3n^{2}-n=1001 \times 2$

$3n^{2}-n=2002$

$3n^{2}-n-2002=0$

Solving this using quadratic equation, we get:

$= \frac{-(-1)+-\sqrt{(-1)^{2}-4 \times 3 \times (-2002)}}{2 \times 3}$

$= \frac{1+-\sqrt{1+24024}}{6}$

$= \frac{1+-\sqrt{24025}}{6}$

$= \frac{1+-155}{6}$

$= \frac{1+155}{6}$ and $\frac{1-155}{6}$

$= \frac{156}{6}$ and $\frac{-154}{6}$

So, the value of $n$ is $26$ and $\frac{-77}{3}$ .

But $n$ is a natural number. Thus, the answer is: $n=\boxed{26}$

Use formula for nth term as: nth term = (3n-1)(n)/2

from the prev.problem we know that no of dots in the n th pentagonal figure is (n/2)(3n-1).
therefore (n/2)(3n-1)=1001;
or, n(3n-1)=2002=26 (3 .26-1)
i.e.,n=26
is the ans.

$$ (½)x n x (3n−1) = 1001 $$ => 3n² − n − 2002 = 0 $$ => n = 26, −77/3 $$ take n = 26