In an A.P., `(S_(p))/(S_(q)) = (p^(2))/(q^(2)), p ne q`, then `(T_(6))/(T_(21))` is equal to

Options

A. `7//2`
B. `2//7`
C. `11/41`
D. `41//11`

Answer

To solve the problem, we need to find the value of \( \frac{T_6}{T_{21}} \) given that \( \frac{S_p}{S_q} = \frac{p^2}{q^2} \) in an arithmetic progression (A.P.).

### Step 1: Understand the formulas for \( S_p \) and \( S_q \)
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write expressions for \( S_p \) and \( S_q \)
Using the formula for the sums, we can write:
\[ S_p = \frac{p}{2} \left(2a + (p-1)d\right) \]
\[ S_q = \frac{q}{2} \left(2a + (q-1)d\right) \]

### Step 3: Set up the equation from the given condition
From the problem, we have:
\[ \frac{S_p}{S_q} = \frac{p^2}{q^2} \]
Substituting the expressions for \( S_p \) and \( S_q \):
\[ \frac{\frac{p}{2} \left(2a + (p-1)d\right)}{\frac{q}{2} \left(2a + (q-1)d\right)} = \frac{p^2}{q^2} \]
This simplifies to:
\[ \frac{p(2a + (p-1)d)}{q(2a + (q-1)d)} = \frac{p^2}{q^2} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ p(2a + (p-1)d) \cdot q^2 = q(2a + (q-1)d) \cdot p^2 \]

### Step 5: Simplify the equation
By simplifying this equation, we can cancel out common terms and rearrange to isolate \( a \) and \( d \). However, we notice that we can directly find \( T_6 \) and \( T_{21} \).

### Step 6: Write expressions for \( T_6 \) and \( T_{21} \)
The \( n^{th} \) term of an A.P. is given by:
\[ T_n = a + (n-1)d \]
Thus,
\[ T_6 = a + 5d \]
\[ T_{21} = a + 20d \]

### Step 7: Find \( \frac{T_6}{T_{21}} \)
Now, we can write:
\[ \frac{T_6}{T_{21}} = \frac{a + 5d}{a + 20d} \]

### Step 8: Relate \( a + 5d \) and \( a + 20d \) to \( p \) and \( q \)
From our earlier simplifications, we can deduce that:
\[ a + 5d = a + \frac{p-1}{2}d \]
and
\[ a + 20d = a + \frac{q-1}{2}d \]
Thus, we can set \( p - 1 = 10 \) and \( q - 1 = 40 \), leading to \( p = 11 \) and \( q = 41 \).

### Step 9: Substitute values back into the ratio
Now substituting back, we find:
\[ \frac{T_6}{T_{21}} = \frac{11}{41} \]

### Final Answer
Thus, the value of \( \frac{T_6}{T_{21}} \) is:
\[ \frac{T_6}{T_{21}} = \frac{11}{41} \]