Let the first term of a series be T_1=6 and its r^ th term T_r=3 T_r-1+6^r, r=2,3, n. If the sum of the first n terms of this series is 1 5 (n^2-12…

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=3 \mathrm{~T}_{\mathrm{r}-1}+6^{\mathrm{r}}, \mathrm{r}=2,3,4, \ldots \mathrm{n} \\ & \mathrm{T}_2=3 . \mathrm{T}_1+6^2 \\ & \mathrm{~T}_2=3.6+6^2...(1)\end{aligned}$​Tr​=3 Tr−1​+6r,r=2,3,4,…nT2​=3.T1​+62 T2​=3.6+62...(1)​ $\begin{aligned} & \mathrm{T}_3=3 \mathrm{~T}_2+6^3 \\ & \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 \\ & \mathrm{~T}_3=3\left(3.6+6^2\right)+6^3 \\ & \mathrm{~T}_3=3^2 .6+3.6^2+6^3...(2)\end{aligned}$​T3​=3 T2​+63 T3​=3 T2​+63 T3​=3(3.6+62)+63 T3​=32.6+3.62+63...(2)​ $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left[1+\frac{6}{3}+\left(\frac{6}{3}\right)^2+\ldots+\left(\frac{6}{3}\right)^{\mathrm{r}-1}\right] \\ & \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left(1+2+2^2+\ldots+2^{\mathrm{r}-1}\right) \\ & \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} 1 \cdot \frac{\left(1-2^{\mathrm{r}}\right)}{(-1)} \\ & \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} \cdot\left(2^{\mathrm{r}}-1\right) \\ & \mathrm{T}_{\mathrm{r}}=\frac{6 \cdot 3^{\mathrm{r}}}{3} \cdot\left(2^{\mathrm{r}}-1\right)\end{aligned}$​Tr​=3r−1⋅6[1+36​+(36​)2+…+(36​)r−1]Tr​=3r−1⋅6(1+2+22+…+2r−1)Tr​=6⋅3r−11⋅(−1)(1−2r)​Tr​=6⋅3r−1⋅(2r−1)Tr​=36⋅3r​⋅(2r−1)​ $\begin{aligned} & \mathrm{T}_{\mathrm{r}}=2 \cdot\left(6^{\mathrm{r}}-3^{\mathrm{r}}\right) \\ & \mathrm{S}_{\mathrm{n}}=2 \Sigma\left(6^{\mathrm{r}}-3^{\mathrm{r}}\right) \\ & \mathrm{S}_{\mathrm{n}}=2 \cdot\left[\frac{6 .\left(6^{\mathrm{n}}-1\right)}{5}-\frac{3 .\left(3^{\mathrm{n}}-1\right)}{2}\right] \\ & \mathrm{S}_{\mathrm{n}}=2\left[\frac{12\left(6^{\mathrm{n}}-1\right)-15\left(3^{\mathrm{n}}-1\right)}{10}\right] \\ & \mathrm{S}_{\mathrm{n}}=\frac{3}{5}\left[4 \cdot 6^4-5 \cdot 3^{\mathrm{n}}+1\right] \\ & \therefore \mathrm{n}^2-12 \mathrm{n}+39=3 \\ & \mathrm{n}^2-12 \mathrm{n}+36=0 \\ & \mathrm{n}=6\end{aligned}$​Tr​=2⋅(6r−3r)Sn​=2Σ(6r−3r)Sn​=2⋅[56.(6n−1)​−23.(3n−1)​]Sn​=2[1012(6n−1)−15(3n−1)​]Sn​=53​[4⋅64−5⋅3n+1]∴n2−12n+39=3n2−12n+36=0n=6​