Q.Two sequences cannot be in both A.P. and G.P. together.

To determine if a sequence can be both an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.) simultaneously, we use the fundamental definitions of both types of sequences.

Let a sequence be $a_1, a_2, a_3, \dots$.

1. Assume the sequence is both an A.P. and a G.P.

   • If it's an A.P. with common difference $d$: $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$.
   • If it's a G.P. with common ratio $r$: $a_2 = a_1 r$ and $a_3 = a_1 r^2$.

2. Equate the corresponding terms:

   • $a_1 + d = a_1 r \quad (1)$
   • $a_1 + 2d = a_1 r^2 \quad (2)$

3. Consider two cases for $a_1$:

   • Case 1: $a_1 = 0$.

     From (1), $d = 0$. From (2), $2d = 0$, which is consistent.

     If $a_1=0$ and $d=0$, the sequence is $0, 0, 0, \dots$. This is an A.P. with $d=0$. It is also a G.P. (with common ratio $r$ undefined or taken as $1$). Thus, the sequence $0, 0, 0, \dots$ is both an A.P. and a G.P.

   • Case 2: $a_1 \neq 0$.

     From (1), $d = a_1 r - a_1 = a_1(r-1)$.

     Substitute this value of $d$ into (2):

     $a_1 + 2[a_1(r-1)] = a_1 r^2$

     Since $a_1 \neq 0$, we can divide the entire equation by $a_1$:

     $1 + 2(r-1) = r^2$

     $1 + 2r - 2 = r^2$

     $r^2 - 2r + 1 = 0$

     $(r-1)^2 = 0$

     This implies $r=1$.

     Now substitute $r=1$ back into the expression for $d$:

     $d = a_1(1-1) = 0$. …

The question asks us to evaluate the statement: "Two sequences cannot be in both A.P. and G.P. together." This phrasing is a bit unusual; it is almost certainly intended to mean "A sequence cannot be simultaneously an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.)." We will proceed with this interpretation, aiming to determine if such a sequence can exist.

To understand this, we need to recall the fundamental definitions of A.P. and G.P. and then explore if a sequence can satisfy both sets of conditions simultaneously.

Concept and Intuition

An Arithmetic Progression (A.P.) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, $d$. For any three consecutive terms $a, b, c$ in an A.P., we have $b - a = c - b$, which simplifies to $2b = a + c$.

A Geometric Progression (G.P.) is a sequence where the ratio between consecutive terms is constant. This constant ratio is called the common ratio, $r$. For any three consecutive terms $a, b, c$ in a G.P., we have $b/a = c/b$ (assuming $a \neq 0, b \neq 0$), which simplifies to $b^2 = ac$.

The core idea is to assume a sequence is both an A.P. and a G.P., and then use the defining properties of both to see what constraints this imposes on the terms of the sequence. If these constraints lead to a contradiction, then no such sequence exists. If they lead to a specific type of sequence, then such sequences are the only ones that can be both.

Step-by-Step Derivation

1. Set up the conditions for a sequence to be both A.P. and G.P. Let the first three terms of a sequence be $a_1, a_2, a_3$. If this sequence is an A.P., then the middle term is the arithmetic mean of its neighbours:

If this sequence is a G.P., then the middle term is the geometric mean of its neighbours (assuming terms are non-zero):

For a sequence to be both an A.P. and a G.P., both conditions $(*)$ and $(**)$ must hold true for any three consecutive terms.

2. Consider the trivial case: The first term is zero.

Suppose $a_1 = 0$.

From the A.P. condition $(*)$: $2a_2 = 0 + a_3 \implies a_3 = 2a_2$.

From the G.P. condition $(**)$: $a_2^2 = 0 \cdot a_3 \implies a_2^2 = 0 \implies a_2 = 0$.

Substituting $a_2 = 0$ into $a_3 = 2a_2$, we get $a_3 = 2(0) = 0$.

This means if the first term is zero, all subsequent terms must also be zero. The sequence would be $0, 0, 0, \dots$.

Let's check this sequence:

* Is $0, 0, 0, \dots$ an A.P.? Yes, with a common difference $d = 0 - 0 = 0$.

* Is $0, 0, 0, \dots$ a G.P.? Yes, as $a_n = a_{n-1}r$ becomes $0 = 0 \cdot r$, which holds for any common ratio $r$. (The ratio $a_2/a_1 = 0/0$ is indeterminate, but the definition $a_n = a_1 r^{n-1}$ holds for $a_1=0$ if all terms are zero).

So, the sequence $0, 0, 0, \dots$ is both an A.P. and a G.P. This is a counterexample to the original statement.

3. Consider the non-trivial case: The first term is non-zero. Assume $a_1 \neq 0$. Since $a_2^2 = a_1 a_3$ and $a_1 \neq 0$, it implies that $a_2$ cannot be zero (otherwise $a_1 a_3 = 0 \implies a_3 = 0$, then from $2a_2 = a_1 + a_3$, we get $0 = a_1 + 0 \implies a_1 = 0$, which contradicts our assumption $a_1 \neq 0$). So, $a_1, a_2, a_3$ must all be non-zero. From the A.P. condition, we can express $a_3$ in terms of $a_1$ and $a_2$:

Substitute this expression for $a_3$ into the G.P. condition:

Expand the right side:

Rearrange the terms to form a quadratic equation:

This is a perfect square trinomial:

Taking the square root of both sides: …