Question

if the 2 , 4 and 7 terms of a non constant AP are in g GP, then the comon ratio of this GP is​

Answer 1

Answer:

The common ratio of the GP is

$\displaystyle{\boxed{\red{\sf\:\dfrac{2}{3}}}}$

Step-by-step-explanation:

We have given that,

The 2ⁿᵈ, 4ᵗʰ and 7ᵗʰ terms of a non constant AP are in GP.

We have to find the common ratio of the GP.

We know that,

A non constant arithmetic progression ( AP ) is an AP in which common difference is not equal to 0, each term is different.

Let the first term of the AP be a.

And the common difference be d.

We know that, nᵗʰ term of an AP is given by,

tₙ = a + ( n - 1 ) * d

⇒ t₂ = a + ( 2 - 1 ) * d

⇒ t₂ = a + 1 * d

⇒ t₂ = a + d

Now,

t₄ = a + ( 4 - 1 ) * d

⇒ t₄ = a + 3d

Now,

t₇ = a + ( 7 - 1 ) * d

⇒ t₇ = a + 6d

Now, t₂, t₄ and t₇ are in GP.

We know that,

In a geometric progression ( GP ), ratio of two consecutive terms is constant.

Let the common ratio of the GP be r.

∴ t₂ : t₄ = t₄ : t₇ = r

$\displaystyle{\implies\sf\:\dfrac{t_2}{t_4}\:=\:\dfrac{t_4}{t_7}\:=\:r}$

$\displaystyle{\implies\sf\:t_2\:\times\:t_7\:=\:t_4\:\times\:t_4}$

$\displaystyle{\implies\sf\:(\:a\:+\:d\:)\:(\:a\:+\:6d\:)\:=\:(\:a\:+\:3d\:)\:(\:a\:+\:3d\:)}$

$\displaystyle{\implies\sf\:a\:(\:a\:+\:6d\:)\:+\:d\:(\:a\:+\:6d\:)\:=\:(\:a\:+\:3d\:)^2}$

$\displaystyle{\implies\sf\:a^2\:+\:6ad\:+\:ad\:+\:6d^2\:=\:a^2\:+\:2\:\times\:a\:\times\:3d\:+\:(\:3d\:)^2}$

$\displaystyle{\implies\sf\:a^2\:+\:7ad\:+\:6d^2\:=\:a^2\:+\:6ad\:+\:9d^2}$

$\displaystyle{\implies\sf\:a^2\:-\:a^2\:+\:7ad\:-\:6ad\:=\:9d^2\:-\:6d^2}$

$\displaystyle{\implies\sf\:0\:+\:ad\:=\:3d^2}$

$\displaystyle{\implies\sf\:ad\:=\:3d^2}$

$\displaystyle{\implies\sf\:a\:=\:\dfrac{3\:d\:\times\:\cancel{d}}{\cancel{d}}}$

$\displaystyle{\implies\:\boxed{\pink{\sf\:a\:=\:3d\:}}}$

Now, we have,

$\displaystyle{\sf\:\dfrac{t_2}{t_4}\:=\:\dfrac{t_4}{t_7}\:=\:r}$

$\displaystyle{\implies\sf\:\dfrac{a\:+\:d}{a\:+\:3d}\:=\:\dfrac{a\:+\:3d}{a\:+\:6d}}$

By substituting a = 3d, we get,

$\displaystyle{\implies\sf\:\dfrac{3d\:+\:d}{3d\:+\:3d}\:=\:\dfrac{3d\:+\:3d}{3d\:+\:6d}}$

$\displaystyle{\implies\sf\:\dfrac{4\:\cancel{d}}{6\:\cancel{d}}\:=\:\dfrac{6\:\cancel{d}}{9\:\cancel{d}}}$

$\displaystyle{\implies\sf\:\dfrac{4}{6}\:=\:\dfrac{6}{9}\:=\:r}$

$\displaystyle{\implies\sf\:r\:=\:\dfrac{\cancel{4}}{\cancel{6}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:r\:=\:\dfrac{2}{3}\:}}}}$

∴ The common ratio of the GP is

$\displaystyle{\boxed{\red{\sf\:\dfrac{2}{3}}}}$