#### The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5thterm to the 21stterm, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

According to question

$\\\frac{{{a}_{11}}}{{{a}_{18}}}=\frac{2}{3} \Rightarrow \frac{a+10d}{a+17d}=\frac{2}{3} (Using a\textsubscript{n} = a + (n - 1).d )\\ \\ 3a + 30d = 2a + 34d\\ 3a - 2a + 30d - 34d = 0\\ a - 4d = 0 a = 4d \ldots (1)\\$
$\\Ratio of 5\textsuperscript{th} and 21th term is \\ \frac{{{a}_{5}}}{{{a}_{21}}}=\frac{a+4d}{a+20d} \ldots (2)\\ Put a = 4d in (2) we get =\frac{4d+4d}{4d+20d}=\frac{8d}{24d} \\$

$\frac{a_{5}}{a_{21}} = \frac{1}{3} \\ \text{Ratio of } S_{5} \text{ to } S_{21} \text{ is } \\ \frac{S_{5}}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} \quad \left[\because S_{n} = \frac{n}{2}[2a + (n-1)d]\right] \\ \frac{S_{5}}{S_{21}} = \frac{5[2a + 4d]}{21[2a + 20d]} \\ \text{Put } a = 4d \text{ in equation (3) we get } \\ \frac{S_{5}}{S_{21}} = \frac{5[2(4d) + 4d]}{21[2(4d) + 20d]} = \frac{5[8d + 4d]}{21[8d + 20d]} = \frac{5[12d]}{21[28d]} = \frac{60d}{588d} \\ \frac{S_{5}}{S_{21}} = \frac{5}{49}$