Find the sum of the first 6 terms of the G.P.
`3, 9/2 , (27)/(4) ,(81)/(8)`,…

Answer

To find the sum of the first 6 terms of the given geometric progression (G.P.):

**Step 1: Identify the first term (a) and the common ratio (r).**
- The first term \( a \) is \( 3 \).
- To find the common ratio \( r \), we can use the formula:
\[ r = \frac{T_2}{T_1} = \frac{9/2}{3} = \frac{9}{2} \times \frac{1}{3} = \frac{9}{6} = \frac{3}{2} \]

**Step 2: Use the formula for the sum of the first n terms of a G.P.**
- The formula for the sum \( S_n \) of the first \( n \) terms of a G.P. is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
- Here, \( n = 6 \), \( a = 3 \), and \( r = \frac{3}{2} \).

**Step 3: Substitute the values into the formula.**
- Substitute \( a \), \( r \), and \( n \) into the formula:
\[ S_6 = 3 \frac{\left(\frac{3}{2}\right)^6 - 1}{\frac{3}{2} - 1} \]

**Step 4: Calculate \( \left(\frac{3}{2}\right)^6 \).**
- First, calculate \( \left(\frac{3}{2}\right)^6 \):
\[ \left(\frac{3}{2}\right)^6 = \frac{729}{64} \]

**Step 5: Substitute back into the sum formula.**
- Now substitute \( \frac{729}{64} \) back into the sum formula:
\[ S_6 = 3 \frac{\frac{729}{64} - 1}{\frac{3}{2} - 1} \]

**Step 6: Simplify the denominator.**
- Simplify \( \frac{3}{2} - 1 \):
\[ \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]

**Step 7: Substitute and simplify the sum expression.**
- Substitute back into the equation:
\[ S_6 = 3 \frac{\frac{729}{64} - 1}{\frac{1}{2}} = 3 \cdot 2 \left(\frac{729 - 64}{64}\right) \]
- Calculate \( 729 - 64 = 665 \):
\[ S_6 = 6 \cdot \frac{665}{64} = \frac{3990}{64} \]

**Step 8: Simplify the fraction.**
- Dividing \( 3990 \) by \( 64 \):
\[ S_6 = 62.34375 \]

Thus, the sum of the first 6 terms of the G.P. is \( 62.34375 \).

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