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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Sequences and Series

Let the first term of a series be T_1=6 and its r^ th term T_r=3 T_...

Let the first term of a series be T1=6T_1=6 and its rth r^{\text{th }} term Tr=3Tr−1+6r,r=2,3T_r=3 T_{r-1}+6^r, r=2,3, n\qquad n. If the sum of the first nn terms of this series is 15(n2−12n+39)(4⋅6n−5⋅3n+1)\frac{1}{5}\left(n^2-12 n+39\right)\left(4 \cdot 6^n-5 \cdot 3^n+1\right), then nn is equal to______

Correct answer

0

Step-by-step solution

Tr=3 Tr−1+6r,r=2,3,4,…nT2=3.T1+62 T2=3.6+62...(1)\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}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}_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} Tr=3r−1⋅6[1+63+(63)2+…+(63)r−1]Tr=3r−1⋅6(1+2+22+…+2r−1)Tr=6⋅3r−11⋅(1−2r)(−1)Tr=6⋅3r−1⋅(2r−1)Tr=6⋅3r3⋅(2r−1)\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=2⋅(6r−3r)Sn=2Σ(6r−3r)Sn=2⋅[6.(6n−1)5−3.(3n−1)2]Sn=2[12(6n−1)−15(3n−1)10]Sn=35[4⋅64−5⋅3n+1]∴n2−12n+39=3n2−12n+36=0n=6\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}$
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