The third term of a G.P is 2 and the first term is 48. Find the sum of the first 11 terms
In a geometric progression (G.P.), the nth term is given by the formula:
\[
T_n = a \cdot r^{(n-1)}
\]
where:
- \(T_n\) is the nth term
- \(a\) is the first term
- \(r\) is the common ratio
From the problem, we are given that:
- \(T_3 = 2\) and \(a = 48\)
Substituting these values into the formula, we have:
\[
2 = 48 \cdot r^{(3-1)}
\]
Simplifying the equation,
\[
2 = 48 \cdot r^2
\]
Dividing both sides of the equation by 48,
\[
r^2 = \frac{1}{24}
\]
Taking the square root of both sides,
\[
r = \pm \sqrt{\frac{1}{24}} = \pm \frac{1}{\sqrt{24}} = \pm \frac{1}{\sqrt{4\cdot6}} = \pm \frac{1}{2\sqrt{6}}
\]
Since \(r\) must be positive for a G.P., we can take:
\[
r = \frac{1}{2\sqrt{6}}
\]
Now, we can calculate the sum of the first 11 terms using the formula:
\[
S_n = \frac{a \cdot (r^n - 1)}{r - 1}
\]
Substituting the given values:
\[
S_{11} = \frac{48 \cdot \left(\left(\frac{1}{2\sqrt{6}}\right)^{11} - 1\right)}{\frac{1}{2\sqrt{6}} - 1}
\]
Simplifying,
\[
S_{11} = \frac{48 \cdot \left(\frac{1}{2^{11}\sqrt{6}} - 1\right)}{\frac{1}{2\sqrt{6}} - \frac{2\sqrt{6}}{2\sqrt{6}}}
\]
\[
S_{11} = \frac{48 \cdot \left(\frac{1}{2^{11}\sqrt{6}} - 1\right)}{\frac{1}{2\sqrt{6}} - \frac{2\sqrt{6}}{2\sqrt{6}}}
\]
\[
S_{11} = \frac{48 \cdot \left(\frac{1}{2^{11}\sqrt{6}} - 1\right)}{-\frac{\sqrt{6}}{2\sqrt{6}}}
\]
\[
S_{11} = 48 \cdot \left(-\frac{1}{2^{11}\sqrt{6}} + 1\right) \cdot \left(\frac{2\sqrt{6}}{\sqrt{6}}\right)
\]
\[
S_{11} = 48 \cdot \left(\frac{2\sqrt{6} - 1}{2^{11}\sqrt{6}}\right) \cdot \left(\frac{2\sqrt{6}}{\sqrt{6}}\right)
\]
\[
S_{11} = \boxed{\frac{96(2\sqrt{6} - 1)}{2^{11}}}
\]