Question

If a, b, c are in AP then ${{2}^{ax+1}}$ , ${{2}^{bx+1}}$ , ${{2}^{cx+1}}$ , $x\ne 0$ are in
1. AP
2. GP only when $x>0$
3. GP if $x<0$
4. GP for all $x\ne 0$

Answer 1

We have given that $a,b,c$ are in AP hence we will apply the formula for three equations in AP. After that we will calculate the value of $\dfrac{B}{A}$ and $\dfrac{C}{B}$ and find out their values by common difference then check which option is correct in the above given options.

Complete step-by-step solution:
We know that the succession of numbers formed according to some definite rule is called a sequence, that is it is a function whose domain is the set of real numbers. Also a progression is a sequence whose members follow a specific rule of pattern.
A progression is a list of numbers or items that exhibit a particular pattern is called progression.
Progression is of two types:
i. Arithmetic Progression
ii. Geometric Progression
Arithmetic Progression: A finite or infinite sequence or series ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}$ or ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}+....$ is said to be an arithmetic progression $(A.P)$ ${{a}_{k}}-{{a}_{k-1}}=d$, a constant independent of $k$ for $k=1,2,3,4,.......,n$ or $k=2,3,4,.....$ Hence the constant $d$ is called the common difference of the arithmetic progression.
Geometric Progression: A sequence is called a geometric progression if the ratio of any term to the preceding terms is a constant called common ratio that is a finite sequence ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}$ or ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}+....$ is said to be geometric progression if none of the ${{a}_{n}}$ is zero and that $\dfrac{{{a}_{k+1}}}{{{a}_{k}}}=r$, a constant $\text{(independent of k)}$ for $k=1,2,3,4,...$
The first term and the common ratio of GP are denoted as $a,r$
Then ${{n}^{th}}$term of a GP is given by the formula: ${{a}_{n}}=a{{r}^{n-1}}$
Now according to the question:
Since $a,b,c$ are in AP.
If three equations are in AP then $2b=a+c$
Let $A={{2}^{ax+1}}$, $B={{2}^{bx+1}}$ , $C={{2}^{cx+1}}$
If we calculate the value of $\dfrac{B}{A}$ then we will get:
$\Rightarrow \dfrac{B}{A}=\dfrac{{{2}^{bx+1}}}{{{2}^{ax+1}}}$
$\Rightarrow \dfrac{B}{A}={{2}^{bx+1-ax-1}}$
$\Rightarrow \dfrac{B}{A}={{2}^{bx-ax}}$
$\Rightarrow \dfrac{B}{A}={{2}^{(b-a)x}}$
$b-a=c-b=d$ that is a common difference as $a,b,c$ are in AP.
Hence $\dfrac{B}{A}={{2}^{dx}}$ as $b-a=d$
Now if we find out the value of $\dfrac{C}{B}$ then we will get:
$\Rightarrow \dfrac{C}{B}=\dfrac{{{2}^{cx+1}}}{{{2}^{bx+1}}}$
$\Rightarrow \dfrac{C}{B}={{2}^{cx+1-bx-1}}$
$\Rightarrow \dfrac{C}{B}={{2}^{cx-bx}}$
$\Rightarrow \dfrac{C}{B}={{2}^{(c-b)x}}$
$\Rightarrow \dfrac{C}{B}={{2}^{dx}}$ as $c-b=d$
Therefore $\dfrac{B}{A}=\dfrac{C}{B}$
$\Rightarrow {{B}^{2}}=AC$
We can say that $A,B,C$ are in GP that is ${{2}^{ax+1}}$, ${{2}^{bx+1}}$ , ${{2}^{cx+1}}$ are in GP for all $x$
Hence option $(4)$ is correct.

Note: We must remember that if each term of a geometric progression is multiplied by a nonzero number, then the sequence obtained is also a geometric progression and if we multiply the corresponding terms of two geometric progressions then the sequence obtained is also a geometric progression.