Let $a_m(m=1,2,3,...,p)$ be the possible integral value of $a$ for which the graphs of $f(x)=ax^2 +2bx + c$ and $g(x)=5x^2-3bx-a$ meet at some point for all real values of $b$. Let $t_r=\prod_{m=1}^p(r-a_m)$ and $S_n=\sum_{r=1}^nt_r$ , $n$ belonging to set of all natural numbers.
Then find the minimum possible value of $a$ and the sum of values of $n$ for which $S_n$ vanishes. Also, find the value of $\sum_{r=5}^\infty\frac{1}{t_r}$.
I first thought of solving the two parabolic curves but it would become tedious.
This question is from one of the textbooks for the IITJEE exam. The question had no subsequent details.
P.S. - I have again posted this question despite getting negative feedback. Please enlighten me
Let us consider $f(x)-g(x)=0,$ i.e. $$(a-5)x^2+5bx+a+c=0\tag1$$
We want $(1)$ to have at least one real solution for all $b$.
For $c=-5$ and $a=5$, $(1)$ has a solution $x=0$ for all $b$.
For $c=-5$ and $a\not=5$, $(1)$ does not have any real solutions for $b=0$.
For $c\not=-5$ and $a=5$, $(1)$ does not have any real solutions for $b=0$.
For $c\not=-5$ and $a\not=5$, considering the discriminant of $(1)$, we see that $(5b)^2-4(a-5)(a+c)\ge 0$ holds for all $b$ if and only if $(a-5)(a+c)\le 0$, i.e. $$\min(5,-c)\le a\le \max(5,-c)\tag2$$
Case 1 : $c\le -6$
It follows from $a\not=5$ and $(2)$ that $a=6,7,\cdots, k$ where $k=\lfloor -c\rfloor$.
$$\begin{align}S_n&=\sum_{r=1}^n(r-6)(r-7)\cdots (r-k) \\\\&=\frac{1}{k-4}\sum_{r=1}^{n}\bigg((r-5)(r-6)\cdots (r-k)-(r-6)(r-7)\cdots (r-k-1)\bigg) \\\\&=\frac{(n-5)(n-6)\cdots (n-k)-(-5)(-6)\cdots (-k)}{k-4}\end{align}$$ using the idea of telescoping series.
If $k$ is odd, then there is no positive integer $n$ such that $S_n=0$.
If $k$ is even, then $S_n=0\iff n-5=k\iff n=k+5$.
Also, $\sum_{r=5}^\infty\frac{1}{t_r}$ is undefined since $t_6=0$.
(This answer considers $\sum_{\color{red}r=5}^{\infty}(1/t_r)$ instead of $\sum_{t=5}^{\infty}(1/t_r)$ you wrote.)
Case 2 : $-6\lt c\lt -4$
From $(2)$, the only possible $a$ is $a=5$. $$S_n=\sum_{r=1}^{n}(r-5)=\frac{n(n-9)}{2}$$ So, $S_n=0\iff n=9$, and $\sum_{r=5}^\infty\frac{1}{t_r}$ is undefined since $t_5=0$.
Case 3 : $-4\le c$
It follows from $a\not=5$ and $(2)$ that $a=k, k+1,\cdots, 4$ where $k=\lceil -c\rceil$.
$$\begin{align}S_n&=\sum_{r=1}^n(r-k)(r-k-1)\cdots (r-4) \\\\&=\frac{1}{6-k}\sum_{r=1}^n\bigg((r+1-k)(r-k)\cdots (r-4)-(r-k)(r-k-1)\cdots (r-5)\bigg) \\\\&=\frac{(n+1-k)(n-k)\cdots (n-4)-(1-k)(-k)\cdots (-4)}{6-k}\end{align}$$
Case 3-1 : $k=4$, i.e. $-4\le c\lt -3$
$$S_n=\frac{(n-3)(n-4)-(-3)(-4)}{2}$$ So, $S_n=0\iff n=7$. $$\sum_{r=5}^\infty\frac{1}{t_r}=\sum_{r=5}^\infty\frac{1}{r-4}=+\infty$$
Case 3-2 : $k=3$, i.e. $-3\le c\lt -2$
$$S_n=\frac{(n-2)(n-3)(n-4)-(-2)(-3)(-4)}{3}$$ There is no positive integer $n$ such that $S_n=0$. $$\sum_{r=5}^\infty\frac{1}{t_r}=\sum_{r=5}^\infty\frac{1}{(r-3)(r-4)}=\sum_{r=5}^\infty\bigg(\frac{1}{r-4}-\frac{1}{r-3}\bigg)=1$$
Case 3-3 : $k=2$, i.e. $-2\le c\lt -1$ $$S_n=\frac{(n-1)(n-2)(n-3)(n-4)-(-1)(-2)(-3)(-4)}{4}$$ So, $S_n=0\iff n=5$. $$\sum_{r=5}^\infty\frac{1}{t_r}=\sum_{r=5}^\infty\frac{1}{(r-2)(r-3)(r-4)}=\sum_{r=5}^\infty\bigg(\frac{1/2}{(r-3)(r-4)}-\frac{1/2}{(r-2)(r-3)}\bigg)=\frac 14$$
Case 3-4 : $k\le 1$, i.e. $c\ge -1$
$$S_n=\frac{(n+1-k)(n-k)\cdots (n-4)}{6-k}$$ So, $S_n=0\iff n=1,2,3,4$. $$\small\begin{align}\sum_{r=5}^\infty\frac{1}{t_r}&=\sum_{r=5}^\infty\frac{1}{(r-k)(r-k-1)\cdots (r-4)} \\\\&=\frac{1}{k-4}\sum_{r=5}^{\infty}\bigg(\frac{1}{(r-k)(r-k-1)\cdots (r-3)}-\frac{1}{(r-k-1)(r-k-2)\cdots (r-4)}\bigg) \\\\&=\frac{1}{(4-k)(4-k)!}\end{align}$$
Conclusion :
The minimum possible value of $a$ is $$\begin{cases}6&\text{if $\ \ c\le -6$} \\\\5&\text{if $\ -6\lt c\lt -5$} \\\\\lceil -c\rceil&\text{if $\ -5\le c$}\end{cases}$$
The sum of values of $n$ for which $S_n$ vanishes is
$$\begin{cases}\text{none}&\text{if $\ \ c\le -6$ and $\lfloor -c\rfloor$ is odd} \\\\\lfloor -c\rfloor+5&\text{if $\ \ c\le -6$ and $\lfloor -c\rfloor$ is even} \\\\9&\text{if $\ -6\lt c\lt -4$} \\\\7&\text{if $\ -4\le c\lt -3$} \\\\\text{none}&\text{if $\ -3\le c\lt -2$} \\\\5&\text{if $\ -2\le c\lt -1$} \\\\10&\text{if $\ -1\le c$} \end{cases}$$
and $$\sum_{t=5}^\infty\frac{1}{t_r}=\begin{cases}\text{undefined}&\text{if $\ \ c\lt -4$} \\\\+\infty&\text{if $\ -4\le c\lt -3$} \\\\\bigg(4-\lceil -c\rceil)(4-\lceil -c\rceil)!\bigg)^{-1}&\text{if $\ -3\le c$} \end{cases}$$