© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Problem
We have an arithmetic progression (A.P.) with first term $a_1$ and common difference $d$. The sum of the first 10 terms is compared to the sum of the first $p$ terms as follows:
$ \displaystyle \frac{a_1 + a_2 + \dots + a_{10}}{a_1 + a_2 + \dots + a_p} \;=\; \frac{100}{p^2}, \quad p \neq 10.
$
We need to find the ratio $ \displaystyle \frac{a_{11}}{a_{10}}.$
Step 2: Write the Sum of the First $n$ Terms of an A.P.
The sum of the first $n$ terms of an A.P. is given by:
$ \displaystyle S_n \;=\; \frac{n}{2} \Bigl(2\,a_1 + (n-1)\,d\Bigr).$
Hence,
$ \displaystyle S_{10} \;=\; \frac{10}{2}\bigl(2\,a_1 + 9\,d\bigr) \;=\; 5\bigl(2\,a_1 + 9\,d\bigr).$
$ \displaystyle S_p \;=\; \frac{p}{2}\bigl(2\,a_1 + (p-1)\,d\bigr).$
Step 3: Form the Given Ratio and Simplify
From the problem statement:
$ \displaystyle \frac{S_{10}}{S_p} \;=\; \frac{100}{p^2}.
$
Substitute the expressions for $S_{10}$ and $S_p$:
$ \displaystyle \frac{\frac{10}{2}\bigl(2\,a_1 + 9\,d\bigr)}{\frac{p}{2}\bigl(2\,a_1 + (p-1)\,d\bigr)}
\;=\; \frac{100}{p^2}.
$
This simplifies to:
$ \displaystyle \frac{5\bigl(2\,a_1 + 9\,d\bigr)}{\frac{p}{2}\bigl(2\,a_1 + (p-1)\,d\bigr)}
\;=\; \frac{100}{p^2}.
$
Canceling common factors carefully, we get:
$ \displaystyle \bigl(2\,a_1 + 9\,d\bigr)\,p
\;=\; 10\,\bigl(2\,a_1 + (p-1)\,d\bigr).
$
Step 4: Rearrange and Simplify the Equation
Expand both sides:
$ \displaystyle p(2\,a_1 + 9\,d) \;=\; 10\bigl(2\,a_1 + (p-1)\,d\bigr).
$
Distribute terms:
$ \displaystyle 2\,p\,a_1 + 9\,p\,d
\;=\; 20\,a_1 + 10\,(p-1)\,d.
$
Rearrange to isolate like terms:
$ \displaystyle 9\,p\,d - 10\,(p-1)\,d \;=\; 20\,a_1 - 2\,p\,a_1.
$
Factor out $d$ and $a_1$ appropriately. After simplification, it leads to an expression involving $ \frac{a_1}{d}.$
Step 5: Solve for $ \displaystyle \frac{a_1}{d} $
Following the algebraic manipulation (as shown in the reference solution), one obtains:
$ \displaystyle 9\,p
\;=\; (20 - 2\,p)\,\frac{a_1}{d} \;+\; 10\,(p-1).
$
Further simplification yields:
$ \displaystyle \frac{a_1}{d} \;=\; \frac{1}{2}.
$
Step 6: Express $ \displaystyle \frac{a_{11}}{a_{10}} $
The $n$-th term of an A.P. is given by $ \displaystyle a_n = a_1 + (n-1)\,d.$ Thus,
$ \displaystyle a_{10} \;=\; a_1 + 9\,d,\quad a_{11} \;=\; a_1 + 10\,d.
$
So the required ratio is:
$ \displaystyle \frac{a_{11}}{a_{10}} \;=\; \frac{a_1 + 10\,d}{a_1 + 9\,d}.
$
Step 7: Substitute $ \displaystyle \frac{a_1}{d} = \frac{1}{2} $
Since $ \displaystyle a_1 = \frac{d}{2},$ substituting into the ratio gives:
$ \displaystyle \frac{a_{11}}{a_{10}}
\;=\; \frac{\frac{d}{2} + 10\,d}{\frac{d}{2} + 9\,d}
\;=\; \frac{\frac{1}{2} + 10}{\frac{1}{2} + 9}
\;=\; \frac{10.5}{9.5}
\;=\; \frac{21}{19}.
$
Step 8: Final Answer
The required ratio is $ \displaystyle \frac{21}{19}.$
