If 1,log _9( 3^1 - x + 2 ),log _3lbrack 4 * 3^x - 1 rbrackn are in A.P., then x equals

n n (b) : As 1, (1 / 2) log _3(3^1-x+2), log _3(4 * 3^x-1) in A.P. n => log _3(3^1-x+2)=log _3(4 * 3^x-1)+1 n => 3^1-x+2=(4 * 3^x-1) x 3 n [Because log _3 3=1] n => 3^i-x+2=12 * 3^x-3 n => 3^x[(3^1-x)+2]=12 * 3^2 x-3 * 3^x n (Multiplying 3^x both sides) n => 12 t^2-5 t-3=0[ where =3^x] n => (3 t+1)(4 t-3)=0 => t=-1 / 3, t=3 / 4 n => 3^x=-1 / 3 which is not possible and t=(3 / 4) => 3^x=(3 / 4) n => x log _3 3=log _3 3-log _3 4n nn n(By taking logarithm at the base 3 both sides)nn => x=1-log _3 4n

If 1,\log_{9}( 3^{1 - x} + 2 ),\log_{3}\lbrack 4 \cdot... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& \text { (b) : As } 1, \frac{1}{2} \log _3\left(3^{1-x}+2\right), \log _3\left(4 \cdot 3^x-1\right) \in \text { A.P. } \\\n& \Rightarrow \log _3\left(3^{1-x}+2\right)=\log _3\left(4 \cdot 3^x-1\right)+1 \\\n& \Rightarrow 3^{1-x}+2=\left(4 \cdot 3^x-1\right) \times 3 \\\n& {\left[\because \log _3 3=1\right]} \\\n& \Rightarrow 3^{i-x}+2=12 \cdot 3^x-3 \\\n& \Rightarrow 3^x\left[\left(3^{1-x}\right)+2\right]=12 \cdot 3^{2 x}-3 \cdot 3^x \\\n& \text { (Multiplying } 3^x \text { both sides) } \\\n& \Rightarrow 12 t^2-5 t-3=0\left[\text { where }=3^x\right] \\\n& \Rightarrow(3 t+1)(4 t-3)=0 \Rightarrow t=-1 / 3, t=3 / 4 \\\n& \Rightarrow 3^x=-1 / 3 \text { which is not possible and } t=\frac{3}{4} \Rightarrow 3^x=\frac{3}{4} \\\n& \Rightarrow x \log _3 3=\log _3 3-\log _3 4\n\end{aligned}\n

\n \n(By taking logarithm at the base 3 both sides)\n

\n\Rightarrow x=1-\log _3 4\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

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