In a G.P. having third term 24 and common ratio 2, Second term is equal to

1 answer

In a geometric progression (G.P.), the \(n\)-th term can be expressed as:

\[ T_n = a \cdot r^{n-1} \]

where \(a\) is the first term and \(r\) is the common ratio.

Given:

  • Third term \(T_3 = 24\)
  • Common ratio \(r = 2\)

The third term can be expressed as:

\[ T_3 = a \cdot r^{3-1} = a \cdot r^2 \]

Substituting the values we have:

\[ 24 = a \cdot 2^2 \]

This simplifies to:

\[ 24 = a \cdot 4 \]

Now, solving for \(a\):

\[ a = \frac{24}{4} = 6 \]

Now that we have the first term \(a\), we can find the second term \(T_2\):

\[ T_2 = a \cdot r^{2-1} = a \cdot r = a \cdot 2 \]

Substituting \(a = 6\):

\[ T_2 = 6 \cdot 2 = 12 \]

Thus, the second term is equal to \( \boxed{12} \).