The ratio of the 11th term to the 18th

Solution:

Let a arid d be the first term and common difference of an AP

Given that, $a_{11}: a_{18}=2: 3$

$\Rightarrow$ $\frac{a+10 d}{a+17 d}=\frac{2}{3}$

$\Rightarrow$ $3 a+30 d=2 a+34 d$

$\Rightarrow$  $a=4 d$ $\ldots$ (i)

Now, $a_{5}=a+4 d=4 d+4 d=8 d$   [from Eq. (i)]

and $a_{n+1}=a+20 d=4 d+20 d=24 d$  [from Eq. (i)]

$\therefore \quad a_{5}: a_{21}=8 d: 24 d=1: 3$

Now, sum of the first five terms, $S_{5}=\frac{5}{2}[2 a+(5-1) d]$

$=\frac{5}{2}[2(4 d)+4 d]$ [from Eq. (i)]

$=\frac{5}{2}(8 d+4 d)=\frac{5}{2} \times 12 d=30 d$

and sum of the first 21 terms, $S_{21}=\frac{21}{2}[2 a+(21-1) d]$

$=\frac{21}{2}[2(4 d)+20 d]$ [from Eq. (i)]

$=\frac{21}{2}(28 d)=294 d$

So, ratio of the sum of the first five terms to the sum of the first 21 terms

S5 :S21 = 30 d :294 d = 5:49