Question

Find three consecutive terms in an AP whose sum is _3 and the product of their cubes is 512

Answer 1

Three consecutive term are $\mathbf{(1-\sqrt{7} \dot{i}),1,(1+\sqrt{7} \dot{i})}$

Step-by-step explanation:

1. Let three term in A.P are

  First term = a-d

  Second term =a

  Third term = a+d

2. Now from given data, sum of three consecutive term are 3.

  So

   First term +Second term+Third term = 3

   a-d + a + a+d = 3

  So

  a= 1          

3. Again from given data, product of cube of three consecutive term are 512.

   $\mathbf{(First term)^{3}\times (Second term)^{3}\times (Third term)^{3}=512 }$

  $\mathbf{(a-d)^{3}\times (a)^{3}\times (a+d)^{3}=512 }$    ...1)

 

4. On putting the value of a in equation 1)

    $\mathbf{(1-d)^{3}\times (1)^{3}\times (1+d)^{3}=512 }$    ...2)

   

   Equation 2) can be written as

   $\mathbf{(1-d^{2})^{3}=512 }$       ...3)

5. On solving equation 3), we get

  $\mathbf{1-d^{2}=8 }$

 Means

 $\mathbf{d^{2}=-7 }$        ...4)

6. From equation 4), it is clear that d is a complex number because square of d is a negative number.

  So

  $\mathbf{d=\sqrt{7} \dot{i} }$  

7. So

    $\mathbf{First term=1-\sqrt{7} \dot{i} }$

    $\mathbf{Second term=1 }$

    $\mathbf{Third term=1+\sqrt{7} \dot{i} }$

 

 

 

 

Answer 2

Answer:

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