If `S_(n)` denotes the sum of n terms of a G.P., prove that: `(S_(10)-S_(20))^(2)=S_(10)(S_(30)-S_(20))`

Answer

To prove that \((S_{10} - S_{20})^2 = S_{10}(S_{30} - S_{20})\), we will start by using the formula for the sum of the first \(n\) terms of a geometric progression (G.P.). The formula for the sum of the first \(n\) terms \(S_n\) of a G.P. is given by:

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

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

### Step 1: Calculate \(S_{10}\) and \(S_{20}\)

Using the formula, we can express \(S_{10}\) and \(S_{20}\):

\[ S_{10} = \frac{a(r^{10} - 1)}{r - 1} \]
\[ S_{20} = \frac{a(r^{20} - 1)}{r - 1} \]

### Step 2: Calculate \(S_{10} - S_{20}\)

Now, we find \(S_{10} - S_{20}\):

\[ S_{10} - S_{20} = \frac{a(r^{10} - 1)}{r - 1} - \frac{a(r^{20} - 1)}{r - 1} \]

Combining the fractions:

\[ S_{10} - S_{20} = \frac{a[(r^{10} - 1) - (r^{20} - 1)]}{r - 1} = \frac{a(r^{10} - r^{20})}{r - 1} \]

Factoring out \(r^{10}\):

\[ S_{10} - S_{20} = \frac{a r^{10}(1 - r^{10})}{r - 1} \]

### Step 3: Calculate \((S_{10} - S_{20})^2\)

Now, we square the expression:

\[ (S_{10} - S_{20})^2 = \left(\frac{a r^{10}(1 - r^{10})}{r - 1}\right)^2 = \frac{a^2 r^{20}(1 - r^{10})^2}{(r - 1)^2} \]

### Step 4: Calculate \(S_{30}\)

Next, we calculate \(S_{30}\):

\[ S_{30} = \frac{a(r^{30} - 1)}{r - 1} \]

### Step 5: Calculate \(S_{30} - S_{20}\)

Now, we find \(S_{30} - S_{20}\):

\[ S_{30} - S_{20} = \frac{a(r^{30} - 1)}{r - 1} - \frac{a(r^{20} - 1)}{r - 1} \]

Combining the fractions:

\[ S_{30} - S_{20} = \frac{a[(r^{30} - 1) - (r^{20} - 1)]}{r - 1} = \frac{a(r^{30} - r^{20})}{r - 1} \]

Factoring out \(r^{20}\):

\[ S_{30} - S_{20} = \frac{a r^{20}(r^{10} - 1)}{r - 1} \]

### Step 6: Calculate \(S_{10}(S_{30} - S_{20})\)

Now we calculate \(S_{10}(S_{30} - S_{20})\):

\[ S_{10}(S_{30} - S_{20}) = \left(\frac{a(r^{10} - 1)}{r - 1}\right) \left(\frac{a r^{20}(r^{10} - 1)}{r - 1}\right) \]

Combining the terms:

\[ S_{10}(S_{30} - S_{20}) = \frac{a^2 r^{20}(r^{10} - 1)^2}{(r - 1)^2} \]

### Step 7: Compare LHS and RHS

We have:

\[ (S_{10} - S_{20})^2 = \frac{a^2 r^{20}(1 - r^{10})^2}{(r - 1)^2} \]
\[ S_{10}(S_{30} - S_{20}) = \frac{a^2 r^{20}(r^{10} - 1)^2}{(r - 1)^2} \]

Since \((1 - r^{10})^2 = (r^{10} - 1)^2\), we conclude that:

\[ (S_{10} - S_{20})^2 = S_{10}(S_{30} - S_{20}) \]

Thus, we have proved the required identity.