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

Hint: 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.