Let `a_(1) , a_(2) , a_(3) , .....` be terms of an A.P. If `(a_(1)+a_(2)+.....+a_(p))/(a_(1)+a_(2)+...... +a_(q))=(p^(2))/(q^(2)) (p ne q)` then find `(a_(6))/(a_(21))` .

Answer

To solve the problem, we need to find the ratio \(\frac{a_6}{a_{21}}\) given the condition involving the sums of terms in an arithmetic progression (A.P.).

Let's denote the first term of the A.P. as \(a_1\) and the common difference as \(d\). The \(n\)-th term of an A.P. can be expressed as:

\[ a_n = a_1 + (n-1)d \]

### Step 1: Write the sums of the first \(p\) and \(q\) terms.

The sum of the first \(n\) terms of an A.P. is given by the formula:

\[ S_n = \frac{n}{2} \times (2a_1 + (n-1)d) \]

So, we can write:

\[ S_p = \frac{p}{2} \times (2a_1 + (p-1)d) \]
\[ S_q = \frac{q}{2} \times (2a_1 + (q-1)d) \]

### Step 2: Set up the equation based on the problem statement.

According to the problem, we have:

\[ \frac{S_p}{S_q} = \frac{p^2}{q^2} \]

Substituting the expressions for \(S_p\) and \(S_q\):

\[ \frac{\frac{p}{2} (2a_1 + (p-1)d)}{\frac{q}{2} (2a_1 + (q-1)d)} = \frac{p^2}{q^2} \]

This simplifies to:

\[ \frac{p(2a_1 + (p-1)d)}{q(2a_1 + (q-1)d)} = \frac{p^2}{q^2} \]

### Step 3: Cross-multiply to eliminate the fractions.

Cross-multiplying gives us:

\[ p(2a_1 + (p-1)d) \cdot q^2 = q(2a_1 + (q-1)d) \cdot p^2 \]

### Step 4: Expand and rearrange the equation.

Expanding both sides, we get:

\[ pq^2(2a_1 + (p-1)d) = qp^2(2a_1 + (q-1)d) \]

This can be rearranged to isolate terms involving \(a_1\) and \(d\):

\[ 2pq^2 a_1 + pq^2(p-1)d = 2qp^2 a_1 + qp^2(q-1)d \]

### Step 5: Collect like terms.

Rearranging gives:

\[ (2pq^2 - 2qp^2)a_1 = (qp^2(q-1) - pq^2(p-1))d \]

Factoring out common terms:

\[ 2pq(p - q)a_1 = (qp^2(q-1) - pq^2(p-1))d \]

### Step 6: Solve for the ratio \(\frac{a_6}{a_{21}}\).

Now, we need to find the ratio of the 6th term to the 21st term:

\[ a_6 = a_1 + 5d \]
\[ a_{21} = a_1 + 20d \]

Thus, we have:

\[ \frac{a_6}{a_{21}} = \frac{a_1 + 5d}{a_1 + 20d} \]

### Step 7: Substitute \(d\) in terms of \(a_1\).

From the earlier derived equation, we can express \(d\) in terms of \(a_1\) and substitute it back into the ratio. However, we can also directly compute the ratio using the values derived from the conditions.

### Final Calculation:

After simplification, we find that:

\[ \frac{a_6}{a_{21}} = \frac{11}{41} \]

Thus, the final answer is:

\[ \frac{a_6}{a_{21}} = \frac{11}{41} \]