If the terms $a,b,c$ are in A.P, then ${{2}^{ax+1}},{{2}^{bx+1}},{{2}^{cx+1}},x\ne 0$ are in
A. A.P
B. G.P only when $x>0$
C. G.P if $x<0$
D. G.P for all $x\ne 0$
Answer
554.7k+ views
Hint: To find the relation among the given number we will use the relation given between $a,b,c$. As we know in an A.P the difference between consecutive terms is the same so we will use this property and get a relation between $a,b,c$. Then we will assume that the given terms ${{2}^{ax+1}},{{2}^{bx+1}},{{2}^{cx+1}},x\ne 0$ are in G.P. Then we will prove our assumption by checking whether the terms satisfy the condition of a G.P and get our desired answer.
Complete step-by-step solution:
It is given to us that $a,b,c$ is an A.P.
We know difference between consecutive terms is same so,
$\begin{align}
& b-a=c-b \\
& \Rightarrow b+b=a+c \\
\end{align}$
$2b=a+c$….$\left( 1 \right)$
Now as we have to find relation between ${{2}^{ax+1}},{{2}^{2bx+1}},{{2}^{cx+1}},x\ne 0$,
Let us assume these are in G.P so for that they should satisfy equation of G.P which states that:
${{B}^{2}}=AC$….$\left( 2 \right)$
Substitute the following in equation (2):
$\begin{align}
& A={{2}^{ax+1}} \\
& B={{2}^{bx+1}} \\
& C={{2}^{cx+1}} \\
\end{align}$
So we get,
$\begin{align}
& {{\left( {{2}^{bx+1}} \right)}^{2}}={{2}^{ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{2bx+2}}={{2}^{ax+1}}\times {{2}^{cx+1}} \\
\end{align}$
Put value from equation (1) on left hand side of above value we get,
$\begin{align}
& {{2}^{\left( a+c \right)x+2}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{ax+cx+2}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{ax+1}}\times {{2}^{cx+1}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
\end{align}$
So we get the right hand side equal to the left hand side so our assumption is right.
So the terms ${{2}^{ax+1}},{{2}^{2bx+1}},{{2}^{cx+1}}$ are in G.P for all $x\ne 0$
Hence correct option is (D).
Note: A progression is a sequence of numbers that follow a specific pattern. A.P is also known as Arithmetic progression which means that the difference between consecutive terms of a sequence is constant. G.P is also known as Geometric Progression; it is a sequence where each succeeding term is equal to the multiplication of preceding term by a fixed number which is also known as common ratio.
Complete step-by-step solution:
It is given to us that $a,b,c$ is an A.P.
We know difference between consecutive terms is same so,
$\begin{align}
& b-a=c-b \\
& \Rightarrow b+b=a+c \\
\end{align}$
$2b=a+c$….$\left( 1 \right)$
Now as we have to find relation between ${{2}^{ax+1}},{{2}^{2bx+1}},{{2}^{cx+1}},x\ne 0$,
Let us assume these are in G.P so for that they should satisfy equation of G.P which states that:
${{B}^{2}}=AC$….$\left( 2 \right)$
Substitute the following in equation (2):
$\begin{align}
& A={{2}^{ax+1}} \\
& B={{2}^{bx+1}} \\
& C={{2}^{cx+1}} \\
\end{align}$
So we get,
$\begin{align}
& {{\left( {{2}^{bx+1}} \right)}^{2}}={{2}^{ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{2bx+2}}={{2}^{ax+1}}\times {{2}^{cx+1}} \\
\end{align}$
Put value from equation (1) on left hand side of above value we get,
$\begin{align}
& {{2}^{\left( a+c \right)x+2}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{ax+cx+2}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
& \Rightarrow {{2}^{ax+1}}\times {{2}^{cx+1}}={{2}^{2ax+1}}\times {{2}^{cx+1}} \\
\end{align}$
So we get the right hand side equal to the left hand side so our assumption is right.
So the terms ${{2}^{ax+1}},{{2}^{2bx+1}},{{2}^{cx+1}}$ are in G.P for all $x\ne 0$
Hence correct option is (D).
Note: A progression is a sequence of numbers that follow a specific pattern. A.P is also known as Arithmetic progression which means that the difference between consecutive terms of a sequence is constant. G.P is also known as Geometric Progression; it is a sequence where each succeeding term is equal to the multiplication of preceding term by a fixed number which is also known as common ratio.
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