Question

If the $2nd5th$ and $9th$ terms of a non-cinsatant $A.P$. are inn $G.P$. then the common ratio of this $G.P$ is

• A. $\frac{8}{5}$
• B. $\frac{4}{3}$
• C. $1$
• D. $\frac{7}{4}$

Answer 1

The correct option is B $\frac{4}{3}$

Second term of an AP can be written as: a + d

Fifth term of an AP can be written as : a + 4d

Ninth term of an AP can be written as: a + 8d

General term of an AP = a + ( n - 1 ) d

According to the question, it is given that the above terms are in GP. We know the relation that:

$⟹b=\sqrt{ac}$ [ Geometric Mean ]

$⟹b^2=ac$

Now considering 2nd term as a, 5th term as b and 9th term as c, we get:

$⟹(a+4d{)}^2=(a+d\left)\right(a+8d)$

$⟹(a^2+8ad+16d^2)=(a^2+8ad+ad+8d^2)$

Bringing all the 'a' terms on LHS and 'd' terms to the RHS, we get:

$⟹(a^2-a^2+8ad-8ad-ad)=8d^2-16d^2$

$⟹-ad=-8d^2$

$⟹ad=8d\left(d\right)$

Cancelling out 'd' from both the sides we get:

$⟹a=8d$

Now we know the relation between common ratio. It is the ratio between 2nd term, 5th term and 9th term. Substituting the values we get:

$⟹$ 2nd term = $8d+d=9d$

$⟹$ 5th term = $8d+4d=12d$

$⟹$ 9th term = $8d+8d=16d$

Therefore new GP =$9d,12d,16d$

Common ratio between the above terms is given as:

$⟹r=\frac{Secondterm}{FirstTerm}$

$⟹r=\frac{12d}{9d}$

$⟹r=\frac{4}{3}$

Hence common ratio is $\frac{4}{3}$.