Question

Using arithmetic mean write seven rational numbers between 12/5 and 10/3.

Answer 1

11 12 only that’s I know any way bye dude

Answer 2

Step-by-step explanation:

Seven rational numbers between $\frac{12}{5}$ and $\frac{10}{3}$ are $\frac{151}{60},\frac{79}{30},\frac{55}{20},\frac{43}{15},\frac{179}{60},\frac{31}{10},\frac{193}{60}$

To find : Using arithmetic mean write seven rational numbers between $\frac{12}{5}$ and $\frac{10}{3}$ ?

Solution :

Using arithmetic mean between $\frac{12}{5}$ and $\frac{10}{3}$

Here,

First term is  $a=\frac{12}{5}$

Last term is $l=\frac{10}{3}$

Number of terms n=9

The last term formula is $l=a+(n-1)d$

$\frac{10}{3}=\frac{12}{5}+(9-1)d$

$\frac{10}{3}-\frac{12}{5}=8d$

$\frac{50-36}{15}=8d$

$d=\frac{14}{15\times 8}$

$d=\frac{7}{60}$

So, the seven rational rational number are

$a_2=a+d$

$a_2=\frac{12}{5}+\frac{7}{60}$

$a_2=\frac{144+7}{60}$

$a_2=\frac{151}{60}$

$a_3=a+2d$

$a_3=\frac{12}{5}+2(\frac{7}{60})$

$a_3=\frac{12}{5}+\frac{7}{30}$

$a_3=\frac{79}{30}$

$a_4=a+3d$

$a_4=\frac{12}{5}+3(\frac{7}{60})$

$a_4=\frac{12}{5}+\frac{7}{20}$

$a_4=\frac{55}{20}$

$a_5=a+4d$

$a_5=\frac{12}{5}+4(\frac{7}{60})$

$a_5=\frac{12}{5}+\frac{7}{15}$

$a_5=\frac{43}{15}$

$a_6=a+5d$

$a_6=\frac{12}{5}+5(\frac{7}{60})$

$a_6=\frac{12}{5}+\frac{35}{60}$

$a_6=\frac{179}{60}$

$a_7=a+6d$

$a_7=\frac{12}{5}+6(\frac{7}{60})$

$a_7=\frac{12}{5}+\frac{7}{10}$

$a_7=\frac{31}{10}$

$a_8=a+7d$

$a_8=\frac{12}{5}+7(\frac{7}{60})$

$a_8=\frac{12}{5}+\frac{49}{60}$

$a_8=\frac{193}{60}$

Therefore, seven rational numbers between $\frac{12}{5}$ and $\frac{10}{3}$ are $\frac{151}{60},\frac{79}{30},\frac{55}{20},\frac{43}{15},\frac{179}{60},\frac{31}{10},\frac{193}{60}$