If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
Question:

If $18^{\text {th }}$ and $11^{\text {th }}$ term of an A.P. are in the ratio $3: 2$, then its $21^{\text {st }}$ and $5^{\text {th }}$ terms are in the ratio

(a) 3 : 2

(b) 3 : 1

(c) 1 : 3

(d) 2 : 3

Solution:

In the given problem, we are given an A.P whose 18th and 11th term are in the ratio 3:2

We need to find the ratio of its 21st and 5th terms

Now, using the formula

$a_{n}=a+(n-1) d$

Where,

$a=$ first tem of the A.P

$n=$ number of terms

$d=$ common difference of the A.P

 

So,

$a_{18}=a+(18-1) d$

 

$a_{18}=a+17 d$

Also,

$a_{11}=a+(11-1) d$

 

$a_{11}=a+10 d$

Thus,

$\frac{a_{18}}{a_{11}}=\frac{3}{2}$

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

$2(a+17 d)=3(a+10 d)$

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

Further solving for a, we get

$34 d-30 d=3 a-2 a$

$4 d=a$ …..(1)

Now,

$a_{21}=a+(21-1) d$

 

$a_{21}=a+20 d$

Also,

$a_{5}=a+(5-1) d$

 

$a_{5}=a+4 d$

So,

$\frac{a_{21}}{a_{5}}=\frac{a+20 d}{a+4 d}$

Using (1) in the above equation, we get

$\frac{a_{21}}{a_{5}}=\frac{4 d+20 d}{4 d+4 d}$

$\frac{a_{21}}{a_{5}}=\frac{24 d}{8 d}$

 

$\frac{a_{21}}{a_{5}}=\frac{3}{1}$

Thus, the ratio of the $21^{\text {st }}$ and $5^{\text {th }}$ term is $3: 1$

Therefore the correct option is (b).

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