(i) Given a G.P. with a=729 and 7th term 64 find `S_(7)`.
(ii) If `{a_(n)}` is a G.P. and `a_(1)=4,r=5`, find `a_(6)` and `S_(6)`.

Answer

Let's solve the problem step by step.

### Part (i)

**Given:**
- First term \( a = 729 \)
- 7th term \( a_7 = 64 \)

**Step 1: Find the common ratio \( r \)**

The formula for the \( n \)-th term of a geometric progression (G.P.) is given by:

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

For the 7th term:

\[ a_7 = a \cdot r^{7-1} = a \cdot r^6 \]

Substituting the known values:

\[ 64 = 729 \cdot r^6 \]

**Step 2: Solve for \( r^6 \)**

Rearranging the equation gives:

\[ r^6 = \frac{64}{729} \]

**Step 3: Simplify \( \frac{64}{729} \)**

We can express \( 64 \) and \( 729 \) as powers:

\[ 64 = 2^6 \quad \text{and} \quad 729 = 3^6 \]

Thus, we have:

\[ r^6 = \left(\frac{2}{3}\right)^6 \]

Taking the sixth root:

\[ r = \frac{2}{3} \]

**Step 4: Find the sum of the first 7 terms \( S_7 \)**

The formula for the sum of the first \( n \) terms of a G.P. is:

\[ S_n = a \frac{1 - r^n}{1 - r} \]

For \( n = 7 \):

\[ S_7 = 729 \cdot \frac{1 - \left(\frac{2}{3}\right)^7}{1 - \frac{2}{3}} \]

**Step 5: Calculate \( S_7 \)**

First, calculate \( 1 - \frac{2}{3} = \frac{1}{3} \):

\[ S_7 = 729 \cdot \frac{1 - \left(\frac{2}{3}\right)^7}{\frac{1}{3}} = 729 \cdot 3 \cdot \left(1 - \left(\frac{2}{3}\right)^7\right) \]

Now calculate \( \left(\frac{2}{3}\right)^7 = \frac{128}{2187} \):

\[ S_7 = 729 \cdot 3 \cdot \left(1 - \frac{128}{2187}\right) \]

Calculating \( 1 - \frac{128}{2187} = \frac{2187 - 128}{2187} = \frac{2059}{2187} \):

\[ S_7 = 729 \cdot 3 \cdot \frac{2059}{2187} \]

Now calculate \( 729 \cdot 3 = 2187 \):

\[ S_7 = 2187 \cdot \frac{2059}{2187} = 2059 \]

**Final answer for Part (i):** \( S_7 = 2059 \)

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### Part (ii)

**Given:**
- First term \( a_1 = 4 \)
- Common ratio \( r = 5 \)

**Step 1: Find the 6th term \( a_6 \)**

Using the formula for the \( n \)-th term:

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

For \( n = 6 \):

\[ a_6 = 4 \cdot 5^{6-1} = 4 \cdot 5^5 \]

Calculating \( 5^5 = 3125 \):

\[ a_6 = 4 \cdot 3125 = 12500 \]

**Step 2: Find the sum of the first 6 terms \( S_6 \)**

Using the sum formula for a G.P. where \( r > 1 \):

\[ S_n = a \frac{r^n - 1}{r - 1} \]

For \( n = 6 \):

\[ S_6 = 4 \cdot \frac{5^6 - 1}{5 - 1} \]

Calculating \( 5^6 = 15625 \):

\[ S_6 = 4 \cdot \frac{15625 - 1}{4} = 15624 \]

**Final answer for Part (ii):**
- \( a_6 = 12500 \)
- \( S_6 = 15624 \)

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