Question

If 18 and 11 term of an

a.p. are in the ratio 3 : 2, then its 21and 5 terms are in the ratio

Answer 1

Hey There,

Let the AP be a, a+d, a+2d, ...


The $n^{th}$ term is given by :

$T_n=a+(n-1)d$


Now, first of all, ratio of 18th and 11th term is 3:2. So,


$\frac{T_{18}}{T_{11}}=\frac{3}{2} \\ \\ \\ \implies \frac{a+17d}{a+10d}=\frac{3}{2} \\ \\ \\ \implies 2\times (a+17d) = 3\times (a+10d) \\ \\ \\ \implies 2a+34d = 3a+30d \\ \\ \\ \implies \boxed{a=4d}$

We will use this relation a=4d.


Now, we need the ratio os 21st and 5th term.


$Ans = \frac{T_{21}}{T_5} \\ \\ \\ \implies Ans = \frac{a+20d}{a+4d} \\ \\ \\ \implies Ans = \frac{4d+20d}{4d+4d} \\ \\ \\ \implies Ans = \frac{24d}{8d} \\ \\ \\ \implies \boxed{Ans=3}$


Hope it helps
Purva
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