Question from JEE Question bank: If $a_1 = 50$ and $a_1 + a_2 + a_3 + .... + a_n = n^{2}a_n \;\;\forall \;n \geq 1$ then $a_{100}$ equals:
a)$1/100$
b)$1/101$
c)$1/50$
d)$1/51$
Here, the correct answer is b) $1/101$. It's a fairly simple problem that can be solved by just substituting in the formula:
$S_n = \frac{n}{2}[a + l]$, where $a$ is the first term, and $l$ is the last term.
So, $S_{100} = 100^{2}a_{100} = \frac{100}{2}[50 + a_{100}]$
$100a_{100} = \frac{1}{2}[50 + a_{100}]$
$200a_{100} = 50 + a_{100}$
$199a_{100} = 50$
$a_{100} = 50/199$
However this doesn't match with the correct answer, b) $1/101$. I presume I've made a very obvious and silly error somewhere early on in my concept, so if someone could point out that error, and guide me to the correct answer, it would be much appreciated.
Edit: Thanks to user Gregory in the comments, I could make out you had to take the difference of Sums $S_n$ and $S_{n-1}$. But I'm still struggling in simplifying it.
$S_{100} = a_1 + a_2 + ..... a_{100} = 100^{2}a_{100}$
$-S_{99} = a_1 + a_2 + ..... a_{99} = 99^{2}a_{99}$
$\implies a_{100} = 100^{2}a_{100} - 99^{2}a_{99}$
But how do we express $a_{99}$ in terms of $a_1$ and $a_{100}$? We cannot use $a + 98d$ as N. F. Tausig pointed out, they haven't mentioned it's an AP.
Edit 2: I followed the advice of users in the comments. Continuing my previous attempt,
$9999a_{100} = 99^{2}a_{99}$
We also get a function for $a_{n}$ by subtracting $S_n$ and $S_{n-1}$ ie.
$a_{n} = \frac{(n-1)^{2}}{n^{2}-1}a_{n-1}$
Hence, $a_{100} = \frac{(99)^{2}}{100^{2}-1}a_{n-1}$
$a_{99} = \frac{(98)^{2}}{99^{2}-1}a_{n-1}$
$a_{98} = \frac{(97)^{2}}{98^{2}-1}a_{n-1}$..... and so on.
We can express each $a_{n}$ in terms of the previous $a_{n}$. This gives us a beautiful solution.
$$a_{100} = \frac{99^2}{100^{2}-1} \times \frac{98^2}{99^{2}-1} \times \frac{97^2}{98^{2}-1} .... \times \frac{4^2}{5^{2}-1} \times \frac{3^2}{4^{2}-1} \times \frac{2^2}{3^{2}-1} \times \frac{1^2}{2^{2}-1}\times a_1$$
$$a_{100} = \frac{99}{101} \times \frac{98}{100} \times \frac{97}{99} .... \times \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 50$$
$$a_{100} = \frac{2 \times 50}{100 \times 101}$$
$$\implies a_{100} = \frac{1}{101}$$
My regards to all the helpful users in the comments who patiently helped me solve a seemingly "elementary" problem for them. I would not have been able to solve this if not for their help.