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Step-by-Step Solution
Step 1: Express the sum of the infinite geometric progression (GP)
The sum of an infinite GP with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by
$ \displaystyle S_{\infty} = \frac{a}{1 - r}. $
According to the question, this sum is 5:
$ \displaystyle \frac{a}{1 - r} = 5. $
Thus,
$ \displaystyle a = 5\,(1 - r). $
Step 2: Use the sum of the first 5 terms of the GP to find $r$
The sum of the first $n$ terms of a GP (for $r \neq 1$) is
$ \displaystyle S_{n} = a \,\frac{1 - r^{\,n}}{1 - r}. $
For $n=5,$ the question states that this sum is $ \frac{98}{25}. $ Hence,
$ \displaystyle a \,\frac{1 - r^{5}}{1 - r} = \frac{98}{25}.
$
Substituting $a = 5\,(1 - r)$ from Step 1, we get:
$ \displaystyle 5\,(1 - r)\,\frac{1 - r^{5}}{1 - r} = \frac{98}{25}
\;\Longrightarrow\; 5\bigl(1 - r^{5}\bigr) = \frac{98}{25}.
$
$ \displaystyle 1 - r^{5} = \frac{98}{125}
\;\Longrightarrow\; r^{5} = 1 - \frac{98}{125} = \frac{27}{125}.
$
Therefore,
$ \displaystyle r = \sqrt[\,5]{\frac{27}{125}} = \left(\frac{3}{5}\right)^{\!\tfrac{3}{5}}.
$
Although it is a less common expression, we keep it in this exact form.
Step 3: Determine $a$
From Step 1,
$ \displaystyle a = 5(1 - r). $
We have $r,$ so in principle we could evaluate $a$ numerically, but we will keep $a$ in symbolic form since the final result is requested in terms of $a_{11}.$
Step 4: Form the arithmetic progression (AP)
According to the question, we now consider an AP whose:
First term $ = 10\,a\,r.$
Common difference $ = 10\,a\,r^{2}.$
General (nth) term $a_{n} = 10\,a\,r + (n-1)\,10\,a\,r^{2}.$
Step 5: Find the 11th term of the AP
The 11th term $a_{11}$ is:
$ \displaystyle a_{11} \;=\; 10\,a\,r + \bigl(11-1\bigr)\,10\,a\,r^{2}
\;=\; 10\,a\,r + 10\cdot 10\,a\,r^{2}
\;=\; 10\,a\,r + 100\,a\,r^{2}.
$
Step 6: Sum of the first 21 terms of the AP
The sum of the first $n$ terms of an AP is given by
$ \displaystyle S_{n} = \frac{n}{2}\,\bigl(2\,A_{1} + (n-1)\,d\bigr), $
where $A_{1}$ is the first term and $d$ is the common difference. For $n = 21,$ we have:
$ \displaystyle S_{21} \;=\; \frac{21}{2}\Bigl[2\,(10\,a\,r) + (21-1)\,(10\,a\,r^{2})\Bigr].
$
$ \displaystyle S_{21} \;=\; \frac{21}{2}\Bigl[20\,a\,r + 200\,a\,r^{2}\Bigr].
$
$ \displaystyle S_{21} \;=\; \frac{21}{2}\,\cdot 20\,a\,r \;+\; \frac{21}{2}\,\cdot 200\,a\,r^{2}
\;=\; 21\bigl(10\,a\,r + 100\,a\,r^{2}\bigr).
$
Notice that
$ \displaystyle \bigl(10\,a\,r + 100\,a\,r^{2}\bigr)
= a_{11}.
$
Hence,
$ \displaystyle S_{21} \;=\; 21\,a_{11}.
$
Final Answer
The sum of the first 21 terms of the given AP is
$ \displaystyle \boxed{21\,a_{11}}.
$
