The ratio of the $11^{t h}$ term to the $18^{t h}$ term of an AP is 2 : 3.
the ratio of the $5^{t h}$ term to the $21^{s t}$ term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms is 1:3; 5:49
if true write 1 and if false then write 0

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Solution

Let $a$ and $d$ be the first term and common difference of given AP. it is given that
$\frac{a_{11}}{a_{18}} = \frac{2}{3}$
$\Rightarrow \frac{a + 10 d}{a + 17 d} = \frac{2}{3}$
$\Rightarrow 3 (a + 10 d) = 2 (a + 17 d)$
$\Rightarrow 3 a + 30 d = 2 a + 34 d$
$\Rightarrow a = 4 d$
Ratio of $5^{t h}$ term to the $21^{s t}$ term is
$\frac{a_{5}}{a_{21}} = \frac{a + 4 d}{a + 20 d}$ ...(1)
putting $a = 4 d$ in (1), we get
$\frac{a_{5}}{a_{21}} = \frac{4 d + 4 d}{4 d + 20 d}$
$\frac{a_{5}}{a_{21}} = \frac{8 d}{24 d}$
$\frac{a_{5}}{a_{21}} = \frac{1}{3}$
Ratio of $S_{5}$ to the $S_{21}$ is
$\frac{S_{5}}{S_{21}} = \frac{\frac{5}{2} [2 a + 4 d]}{\frac{21}{2} [2 a + 20 d]}$ ( $∵ S_{n} = \frac{n}{2} (2 a + (n - 1) d$ )
$\frac{S_{5}}{S_{21}} = \frac{5 [2 a + 4 d]}{21 [2 a + 20 d]}$ ...(2)
putting $a = 4 d$ in (2), we get
$\frac{S_{5}}{S_{21}} = \frac{5 [2 (4 d) + 4 d]}{21 [2 (4 d) + 20 d]}$
$\frac{S_{5}}{S_{21}} = \frac{5 [12 d]}{21 [28 d]}$
$\frac{S_{5}}{S_{21}} = \frac{60 d}{588 d}$
$\frac{S_{5}}{S_{21}} = \frac{5}{49}$

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