Question

The first and third term and seventh term of an ap are consecutive terms of a gp find the common ratio

Answer 1

Let the first term of the arithmetic progression (AP) be \( a \) and the common difference be \( d \). The \( n \)-th term of an AP can be expressed as:

\[
T_n = a + (n-1)d
\]

Given terms are as follows:

- The first term (\( T_1 \)):
\[
T_1 = a
\]

- The third term (\( T_3 \)):
\[
T_3 = a + 2d
\]

- The seventh term (\( T_7 \)):
\[
T_7 = a + 6d
\]

According to the problem, these three terms must form a geometric progression (GP). In a GP, the square of the middle term equals the product of the other two terms. Therefore, we can write the following equation:

\[
(T_3)^2 = T_1 \cdot T_7
\]

Substituting the expressions for the terms we have:

\[
(a + 2d)^2 = a \cdot (a + 6d)
\]

Expanding both sides:

\[
(a^2 + 4ad + 4d^2) = a^2 + 6ad
\]

Now, simplifying this equation:

\[
a^2 + 4ad + 4d^2 = a^2 + 6ad
\]

Subtract \( a^2 \) from both sides:

\[
4ad + 4d^2 = 6ad
\]

Rearranging gives:

\[
4d^2 + 4ad - 6ad = 0
\]

\[
4d^2 - 2ad = 0
\]

Factoring out \( 2d \):

\[
2d(2d - a) = 0
\]

Thus, this gives us two possible scenarios:

1. \( d = 0 \) (which would mean all terms are equal, not forming a progression).
2. \( 2d - a = 0 \) implies \( a = 2d \).

Now we need to find the common ratio \( r \) of the GP. The terms in GP are \( a \), \( a + 2d \), and \( a + 6d \).

Substituting \( a = 2d \):

1. First term:
\[
T_1 = 2d
\]

2. Third term:
\[
T_3 = 2d + 2d = 4d
\]

3. Seventh term:
\[
T_7 = 2d + 6d = 8d
\]

Now, finding the common ratio \( r \):

\[
r = \frac{T_3}{T_1} = \frac{4d}{2d} = 2
\]

And also:

\[
r = \frac{T_7}{T_3} = \frac{8d}{4d} = 2
\]

Therefore, the common ratio of the geometric progression is:

\[
\boxed{2}
\]