If `(a_1+a_2+.....+a_(10))/(a_1+a_2+.....a_p)=(100)/p^2` and `a_i` is in A.P, find value of `(a_(11)/a_(10))`

Answer

To solve the problem, we need to find the value of \( \frac{a_{11}}{a_{10}} \) given that the terms \( a_1, a_2, \ldots, a_{10} \) are in arithmetic progression (A.P.) and the following equation holds:

\[ \frac{a_1 + a_2 + \ldots + a_{10}}{a_1 + a_2 + \ldots + a_p} = \frac{100}{p^2} \]

### Step 1: Express the sums in terms of \( a_1 \) and \( d \)

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

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

For \( n = 10 \):

\[ S_{10} = \frac{10}{2} \left(2a_1 + 9d\right) = 5(2a_1 + 9d) = 10a_1 + 45d \]

For \( n = p \):

\[ S_p = \frac{p}{2} \left(2a_1 + (p-1)d\right) = \frac{p}{2} (2a_1 + (p-1)d) \]

### Step 2: Substitute the sums into the given equation

Substituting \( S_{10} \) and \( S_p \) into the equation:

\[ \frac{10a_1 + 45d}{\frac{p}{2} (2a_1 + (p-1)d)} = \frac{100}{p^2} \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying gives:

\[ (10a_1 + 45d) p^2 = 100 \cdot \frac{p}{2} (2a_1 + (p-1)d) \]

This simplifies to:

\[ (10a_1 + 45d) p^2 = 50p(2a_1 + (p-1)d) \]

### Step 4: Expand both sides

Expanding both sides:

\[ 10a_1 p^2 + 45d p^2 = 100a_1 p + 50p(p-1)d \]

### Step 5: Rearranging the equation

Rearranging gives:

\[ 10a_1 p^2 - 100a_1 p + 45d p^2 - 50p(p-1)d = 0 \]

### Step 6: Factor out common terms

Factoring out \( p \):

\[ p(10a_1 p - 100a_1 + 45d p - 50(p-1)d) = 0 \]

### Step 7: Solve for \( d \)

From the equation, we can isolate \( d \):

\[ 10a_1 p - 100a_1 + 45d p - 50pd + 50d = 0 \]

This leads to:

\[ d(45p - 50p + 50) = 100a_1 - 10a_1 p \]

### Step 8: Find the ratio \( \frac{a_{11}}{a_{10}} \)

Now, we need to find \( \frac{a_{11}}{a_{10}} \):

\[ a_{11} = a_1 + 10d \]
\[ a_{10} = a_1 + 9d \]

Thus,

\[ \frac{a_{11}}{a_{10}} = \frac{a_1 + 10d}{a_1 + 9d} \]

### Step 9: Substitute \( d = 2a_1 \)

From the earlier steps, we found that \( d = 2a_1 \):

\[ \frac{a_{11}}{a_{10}} = \frac{a_1 + 10(2a_1)}{a_1 + 9(2a_1)} = \frac{a_1 + 20a_1}{a_1 + 18a_1} = \frac{21a_1}{19a_1} = \frac{21}{19} \]

### Final Answer

Thus, the value of \( \frac{a_{11}}{a_{10}} \) is:

\[ \frac{21}{19} \]