JEE Main 2024 — Sequences And Series Question with Solution

$$\begin{array}{rl}& T_r=3T_{r-1}+6^r\\ & \Rightarrow \text{ solving homogenous part }\\ & T_r=3T_{r-1}\\ & \Rightarrow x=3\text{ is the root }\end{array}$$

$$\therefore T_r=a.3^r$$

Solving for particular part

$$\begin{array}{rl}& T_r=b.6^r\\ & b.6^r=3b{6}^{r-1}+6^r\\ & \Rightarrow 6b=3b+6\\ & \Rightarrow 3b=6\\ & \Rightarrow b=2\\ & T_r=a^n+a^p\\ & T_r=a{3}^{br}+2.6^r\text{.... (i)}\\ & T_r=3T_{r-1}+6^r\end{array}$$

Putting $$r=2$$

$$T_2=18+36=54\text{.... (ii)}$$

Using equation (i) and (ii)

$$\begin{array}{rl}& 54=9a+72\Rightarrow -18=9a\Rightarrow a=-2\\ & \therefore T_r=2\cdot 6^r-2\cdot 3^r=2\left(6^r-3^r\right)\\ & \sum _{r=1}^n{T}_r=2\sum 6^r-2\sum 3^r\\ & =2\cdot 6\frac{\left(6^n-1\right)}{5}-2\cdot 3\frac{\left(3^n-1\right)}{2}\\ & =\frac{3}{5}\left(4\cdot 6^n-5\cdot 3^n+1\right)\\ & \therefore n^2-12n+39=3\\ & n^2-12n+36=0\\ & (n-6{)}^2=0\\ & \therefore n=6\end{array}$$