Question

if the fourth term of an ap is 0, then a25/a11= please don't scam ​

Answer 1

Answer:

verify the closure property for addition and multiplication for the rational numbers_5/7and8/9

Answer 2

$\dfrac{a {\tiny25}}{a{ \tiny11}} = \dfrac{ 3}{ 1}$

Step-by-step explanation:

★Given -

• Fourth term of an A.P is 0.

★To find -

• $\bf \dfrac{a {\tiny25}}{a{ \tiny11}}$

★Solution -

We know that,

$\bf a{ \tiny4} = 0$

$\longmapsto \bf a + 3d = 0$

$\longmapsto \bf a = - 3d \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

$\mathrm{ Now} \: , \: \bf \dfrac{a {\tiny25}}{a{ \tiny11}} = \dfrac{a + 24d}{a + 10d}$

$\longmapsto\bf \dfrac{a {\tiny25}}{a{ \tiny11}} = \dfrac{ - 3d+ 24d}{ - 3d+ 10d} {\normalsize (from \: \: 1)}$

$\longmapsto\bf \dfrac{a {\tiny25}}{a{ \tiny11}} = \dfrac{ \cancel{21d}}{ \cancel{7d}}$

$\longmapsto\bf \green{\boxed{\dfrac{a {\tiny25}}{a{ \tiny11}} = \dfrac{ 3}{ 1} }}$