Question

if A1,A2 be two arithmetic means between 1/3 and 1/24 their value are​

Answer 1

Step-by-step explanation:

1/3, A1, A2 , 1/24 are in AP

Let A1 = 1/3+d

A2 = 1/3+2d

So 1/24 = 1/3+ 3d

solving it we get d= -7/72

so A1= 17/72

A2 = 10/72 = 5/36

Answer 2

The values are $A_1=\frac{17}{72}$ and $A_2=\frac{5}{36}$.

Step-by-step explanation:

Given : If A1, A2 be two arithmetic means between 1/3 and 1/24.

To find : Their value ?

Solution :

Using the formula  $d=\frac{b-a}{n+1}$

Where, d is the common difference

a and b are the terms in between two arithmetic means lie.

So,  $\frac{1}{3} ,\ A_1,\ A_2 ,\ \frac{1}{24}$ area in A.P.

Let $A_1=\frac{1}{3}+d$

and $A_2=\frac{1}{3}+2d$

Now, $\frac{1}{24} =\frac{1}{3}+ 3d$

Solving the equation,

$3d=\frac{1}{24}-\frac{1}{3}$

$3d=\frac{1-8}{24}$

$3d=\frac{-7}{24}$

$d=\frac{-7}{72}$

So,$A_1=\frac{1}{3}+d=\frac{1}{3}-\frac{7}{72}$

$A_1=\frac{24-7}{72}$

$A_1=\frac{17}{72}$

$A_2=\frac{1}{3}+2d=\frac{1}{3}-2\times \frac{7}{72}$

$A_2=\frac{24-14}{72}$

$A_2=\frac{10}{72}$

$A_2=\frac{5}{36}$

Therefore, the values are $A_1=\frac{17}{72}$ and $A_2=\frac{5}{36}$.

#Learn more

Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is

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