If _ 10 2, _ 10 (2^x-1) and _ 10 (2^x+3) are three consecutive terms of an A.P, then the value of x is

Correct answer: C. _2 5 _ 10 2, _ 10 (2^ x -1) and _ 10 (2^ x +3) are in A.P. Hence, common difference will be same. aligned & _ 10 (2^ x -1)- _ 10 2= (2^ x +3)- _ 10 (2^ x -1) & _ 10 ( 2^ x -1 2 )= _ 10 ( 2^ x +3 2^ x -1 ) & 2^ x -1 2 = 2^x+3 2^ x -1 (2^ x -1)^2=2 (2^ x +3) & 2^ 2 mathrm x -2^ x +1 +1=2^ x+1 +6 aligned 2^ 2 x -2^ x+2 =5 Let 2^x=y , then aligned & y²-4 y-5=0 & y²-5 y+y-5=0 & y(y-5)+1(y-5)=0 & y=-1, y=5 aligned Therefore, 2^ x =5x= _2 5

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Sequences and Series

NDA 2015 · Mathematics · Sequences and Series

If _ 10 2, _ 10 (2^x-1) and _ 10 (2^x+3) are three consecutive...

\log _{10} 2, \log _{10} \left(2^x-1\right)

and

\log _{10} \left(2^x+3\right)

are three consecutive terms of an A.P, then the value of

x

Options

A1
B $\log _5 2$
C $\log _2 5$
D $\log _{10} 5$

Correct answer

C. $\log _2 5$

Step-by-step solution

\log _{10} 2, \log _{10} \left(2^{\mathrm{x}}

-1\right)

and \log _{10} \lleft(2^{\mathrm{x}}

+3)

are in A.P. Hence, common difference will be same.

\begin{aligned} \therefore& \log _{10}\lleft(2^{\mathrm{x}}-1)-\log _{10} 2=\log \l \left(2^{\mathrm{x}}+3\right)-\log _{10}\lleft(2^{\mathrm{x}}-1) \\ \therefore & \log _{10}\left(\frac{2^{\mathrm{x}}-1}{2}\right)=\log _{10}\left(\frac{2^{\mathrm{x}}+3}{2^{\mathrm{x}}-1}\right)\\ \Rightarrow & \frac{2^{\mathrm{x}}-1}{2}=\frac{2^x+3}{2^{\mathrm{x}}-1}\\ \lleft(2^{\mathrm{x}}-1)^2=2\lleft(2^{\mathrm{x}}+3) \\ & 2^{2 \m mathrm{x}}-2^{\mmathrm{x}+1}+1=2^{\mathrm{x}+1}+6\end{aligned}

2^{2 x}-2^{x+2}=5

Let

2^x=y

, then

\begin{aligned} & y²-4 y-5=0 \\ & y²-5 y+y-5=0 \\ & y(y-5)+1(y-5)=0 \\ & y=-1, y=5 \end{aligned}

Therefore, 2^{\mathrm{x}}=5x=\log _2 5

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